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New Chapter 13, Solids I: Theory 13.1 Chapter 13 Solids I: Theory 13.1 Introduction 13.2 Bonding of Solids 13.3 Crystals and Noncrystals 13.4 Energy Levels of Electrons in a Solid; Bands 13.5 Conductors and Insulators 13.6 The Drude Model of Conductivity 13.7 Electron Collisions in Metals 13.8 The Fermi Speed 13.9 Degeneracy Pressure 13.10 White Dwarfs, Neutron Stars, and Black Holes 13.11 Classical and Quantum Gases 13.12 Bose-Einstein Condensation Problems for Chapter 13 Advanced Topics which may be omitted without serious loss of continuity 13.1 Introduction In Chapter 12 we saw how atoms can bond together to form molecules. In this chapter we study the aggregation of atoms on a larger scale, to form solids. There are many close parallels between solid-state and molecular physics. Several of the bonding mechanisms in solids are exactly the same as in molecules; for example, many solids are covalently or ionically bonded. There are, however, many differences between solid-state and molecular physics. For example, the bonding of metallic solids has no direct parallel in molecules. Also, certain solids, instead of being simple aggregates of atoms, are better viewed as being built in two stages, with atoms bound into molecules and then molecules bound into a solid. (A good example is ice, which is best understood as an aggregate of water molecules.) Another new feature of the solid state is the possibility that atoms (or molecules) can bond together into a symmetric, repetitive crystalline array; the study of the possible crystal structures is an important part of solid-state physics with no exact counterpart in molecular physics. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.2 Perhaps the most striking thing about solids is that they were the first example of macroscopic systems that were understood with the help of quantum mechanics.* Numerous commonplace properties of solids — specific heat, conductivity, transparency, magnetic susceptibility, compressibility, and many more — had defied classical explanation, but could be explained and calculated using quantum mechanics. It is this ability to explain — and hence control — the electronic and mechanical properties of solids that has led to a vast array of technological applications which have transformed civilization in the last century. Probably the most important technological advance in solid-state physics was the invention of the transistor, which led to miniturized electronics, personal computers, and the digital information age we now live in. The transistor and several other applications of solid state physics are examined in the next chapter. In this chapter, we concentrate on the theory underlying these applications. The theory of solids necessarily involves approximations. In principle, any property of any solid can be predicted using Schrödinger's equation, but in practice this strategy is fruitless. In the case of a macroscopic solid with some 1023 electrons and nuclei, Schrödinger's equation is exceedingly complex and impossible to solve exactly. (Even in the case of simple molecules with just a few atoms, Schrödinger's equation cannot be solved exactly.) The art of solid state theory is to model the system under study by simply ignoring most of the real-world complexities and concentrate on just a few essential features. For instance, in computing the electronic properties of metal, one often assumes that the conduction electons behave as free electrons in a box — thus the theorist completely ignores the interactions of the electrons with the nuclei and with each other. A theoretical model must be simple enough to allow calculations, but not so simple as to erase the properties the theorist wishes to explain. The check that the model is correct, that its approximations are valid, is always experiment. We begin this chapter, in Sections 13.2 – 13.5, with a broad, qualitative overview of the physics of solids. In Section 13.2 we describe the principal mechanisms by which atoms or molecules bond to one another to form solids. In Section 13.3 we describe some of the regular crystalline structures in which many solids can form, and we contrast these with the irregular, noncrystalline structure of the amorphous solids and the still more irregular composition of liquids and gases. We next take up what is perhaps the single most important achievement of solid-state physics, the theory of electrical conductivity. This theory depends on an understanding of the electron energy levels in a solid. In Section 13.4 we discuss these levels and describe how they fall into bands of allowed energies. Within these bands the energy levels are distributed almost continuously, but between them are gaps where there are no allowed levels. In Section 13.5 we show how the different possible arrangements of these bands explain why some solids are conductors, some are insulators, and some — the so-called semiconductors — are in between. Following our introductory survey of the field, we take a closer look at the use of quantum mechanics in explaining properties of solids, with an emphasis on electronic * The applications of quantum theory to solids go back to the 1907 paper of Einstein, in which he used quantization to explain the specific heats of solids. See Section 15.10. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.3 properties. In Sections 13.6 – 13.8, we examine a relatively simple but very useful model of electrical conduction called the Drude model. We will see that, at almost every step in our discussion of the Drude model, drastic approximations are made in order to keep the theory as simple as possible — and yet the theory agrees very well with experiment and explains many basic properties of metals and semiconductors. We conclude the chapter with some further, somewhat exotic, uses of solid state theory. Quantum mechanics explains why solids are so difficult to compress, compared to gases, and this effect, called degeneracy pressure, is explained in Section 13.10. In Section 13.11, we will see that degeneracy pressure has profound consequences in astrophysics. In the final two sections we describe a very strange state of matter that occurs only at very low temperatures, the Bose-Einstein condensate. 13.2 Bonding of Solids Solids, like molecules, are held together by several different types of bonds. Just as with molecules, the distinctions between the various bonds are not always clear cut, and many solids are bonded by a combination of several mechanisms. Nevertheless, we can distinguish four main types of bonds — covalent, metallic, ionic, and dipole-dipole — and we describe each of these in turn. COVALENT BONDING We saw in Chapter 12 that many molecules are held together by covalent bonds, each comprising two valence electrons shared by two of the atoms. In H2, for example, the two hydrogen atoms share their two electrons to form a single bond; in CH4 (methane), each of the four valence electrons of the carbon joins with an electron from one of the four hydrogens to form a covalent bond. This same mechanism holds the atoms together in many solids. A beautiful example of such a covalent solid is diamond, which is one of the forms of solid carbon. In diamond each carbon atom is covalently bonded to four neighboring carbon atoms. The four bonds from any one carbon atom are directed symmetrically outward, and each atom in diamond is at the center of a regular tetrahedron formed by its four nearest neighbors. [XX Need diagram of tetrahedral diamond structure.] Both silicon and germanium — in the same group as carbon in the periodic table — are bonded covalently in a similar tetrahedral structure. Other examples of covalent solids are silicon carbide (widely used for its hardness and resistance to high temperatures) and zinc sulfide (used by Rutherford and other pioneers in nuclear physics as a scintillation detector). An important characteristic of a solid is its strength. One simple measure of this strength is the atomic cohesive energy, defined as the energy needed, per atom, to separate the solid completely into individual atoms. In Table 13.1 we have listed a number of solids with their cohesive energies. The first three entries are covalent solids, and we see what is generally true for covalent solids, that their cohesive energies vary from around 3 to about 7 eV per atom. This energy scale is not surprising, since chemical bonds involve the outer electrons in atoms, and a few eV is a typical energy for outer- 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.4 shell electrons. As one would expect, those solids with high cohesive energies, like diamond and silicon carbide, are especially strong and have high melting points. [TABLE 13.1XX] Since their electrons are usually well bound and cannot move easily, most covalent solids are poor conductors of electricity. In many covalent solids, the excitation energy of the electrons is higher than the energy of visible photons (2 or 3 eV). For this reason many covalent solids (like diamond) cannot absorb visible photons and are therefore transparent to light. METALLIC BONDS As seen in Figure 13.1, more than half of all elements are metals, and all solid metals are held together by the metallic bond. In some ways, this bond can be seen as a relative of the familiar covalent bond. In a covalent bond, electrons are shared between pairs of atoms. In the metallic bond, electrons are shared among all the atoms: One or two valence electrons are completely detached from each parent atom and can wander freely through the whole metal. Thus a metal can be thought of as a lattice of positive atomic cores, immersed in a "sea" of mobile electrons, as in Figure 13.2. It is the attraction between the positive atomic cores and this negative sea of electrons that holds the metal together. As your can see in Table 13.1, the cohesive energy of typical metals is less than that of many covalent solids, but is, nevertheless, high enough to make most metals rather strong. [FIG 13.1 XX] [FIG 13.2 XX] All metals are good conductors of electricity because the outermost one or two electrons per atom are free to move throughout the entire volume of the metal almost as if they were free electrons. For this reason, they are sometimes described as a "gas" of electrons. These free electrons are called conduction electrons, since they carry, or conduct, electrical current. Since the conduction electrons can also carry energy, metals are good thermal conductors as well. In addition, conduction electrons are the cause of the opaque, but shiny appearance of metal surfaces. When a photon strikes a metal, it can easily lose energy and momentum to the "free" electrons, and metals are therefore not transparent to light. In the language of classical electromagnetism, when visible light strikes the surface of a metal, the oscillating electric field of the light wave exerts an oscillating force on the electrons in the metal. Since the conduction electrons are free to move, they oscillate with large amplitude in response to the field, and their large oscillations create a new electromagnetic wave, the reflected wave, which produces the mirror-like appearance of metals. IONIC BONDS We saw in Chapter 12 that certain pairs of elements can bond ionically to form molecules. For exactly the same reasons, these same pairs of elements can bond ionically 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.5 to form solids. For example, in solid NaCl (common salt) each Na atom has lost one electron to a Cl atom; the resulting closed-shell ions arrange themselves so that each Na+ ion is surrounded by several Cl– ions, and vice versa.* The resulting solid is therefore quite tightly bound. As illustrated by the examples in Table 13.1, typical atomic cohesive energies (energy to separate the solid into neutral atoms) are 3 or 4 eV per atom for ionic solids. Since the electrons of an ionic solid are all bound tightly in closed-shell ions, ionic solids are poor conductors of electricity and heat. For the same reason visible photons cannot be absorbed easily, and ionic solids are transparent to light. DIPOLE-DIPOLE BONDS The ionic bond results from the simple Coulomb force between oppositely charged ions. If the basic units (atoms or molecules) are electrically neutral, this force is not present. There is, nevertheless, a related, though weaker attraction that can act, even between neutral atoms or molecules. A neutral atom or molecule can have its charges distributed so that it has a electric dipole moment; that is, the preponderance of positive charge is centered at one point and that of negative at another. Two neutral dipoles can exert an electrostatic force on one another, as we now describe. Let us consider first two permanent dipoles, that is, two objects each of which is a dipole even when in complete isolation. (An example of a permanent dipole is the water molecule, as we saw in Chapter 12) If we bring these dipoles together so that two like charges are closest, as in Figure 13.3 (a), the repulsive forces are slightly greater than the attractive, and the net force is repulsive (See Problems 13.3XX and 13.4XX). On the other hand, if the positive charge in one dipole is closest to the negative in the other, as in Fig. 13.3(b), the net force is attractive. In addition to these forces, each dipole exerts a torque on the other, and this torque tends to rotate them into the alignment of maximum attraction (c). Therefore, the net force between two permanent dipoles (that are free to rotate) is automatically attractive. Because it involves substantial cancellations between attractive and repulsive forces, this dipole-dipole attraction is much weaker than the Coulomb forces between the individual charges. Nevertheless, it is the attraction that holds many solids together. [FIG 13.3 = old 17.1 XX] Because of their basic spherical symmetry, no atoms are permanent dipoles. On the other hand, many molecules (the so-called polar molecules, like H2O) are, and these molecules can be bound into solids by the dipole-dipole attraction just described. Because the bond holding the molecules to one another is much weaker than the bond holding the atoms together inside each molecule, these solids are properly seen as composed of molecules and are sometimes called molecular solids. An example of a molecular solid is ice; we see from Table 13.1 that the energy needed to separate ice into H2O molecules is * Because each Na+ "belongs" to several Cl– ions, and vice versa, you should not think of solid NaCl as made up of NaCl molecules; rather, it is a bonding of many Na atoms with many Cl atoms , forming one large, crystalline array. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.6 about 0.5 eV per molecule. As expected, this is appreciably less than the cohesive energies of typical covalent, metallic, or ionic solids (usually several eV per atom). It is also much less than the 5 eV needed to remove any one atom from the H2O molecule — which confirms that ice is properly regarded as an assembly of H2O molecules. The bond between two water molecules actually involves another effect as well. We saw in Section 12.5 that the oxygen atom in water appropriates at least part of the electron from each hydrogen, leaving nearly bare protons at the two ends of the molecule. When water molecules bond to form ice, these protons are shared with nearby oxygen atoms in somewhat the same way that electrons are shared in a covalent bond. This bonding by shared protons is called hydrogen bonding and plays an important role in biology: The DNA molecule comprises two intertwined helical strands of atoms. The atoms within each strand are bound together by covalent bonds, while the binding between the two strands is by hydrogen bonds, which are much weaker. This is what allows the two strands to unravel from one another, without disturbing their internal structure, in the process of replication. Although all atoms and many molecules have zero permanent dipole moment, they can acquire an induced moment when put in the field of other charges. This is because the field pushes all positive charges one way and pulls all negatives the other. These induced dipoles are responsible for the very weak van der Waals bond* between some atoms and molecules, as we describe shortly. Because this bond is so weak, it is important only in solids where all other types of bond are absent. For example, the noble gas atoms are subject to none of the bonding mechanisms described earlier, but can form solids because of the van der Waals attraction. Because the bond is so feeble, the noble gases solidify only at rather low temperatures. For instance, as shown in Table 13.1, solid argon has a cohesive energy of only 0.08 eV/atom, and its melting point is 84 K. The van der Waals force is also responsible for the solids of certain molecules that have no permanent dipole moment, such as O2, N2, and CH4 (methane). The origin of the van der Waals force between two atoms (or molecules) that have no permanent dipole moments can be explained by the following classical argument. A classical atom consists of Z electrons orbiting around a fixed nucleus. The statement that it has no permanent dipole moment means (classically) that, when averaged over time, the dipole moment is zero. Nevertheless, as the electrons move around, they can produce an instantaneous nonzero dipole moment. Now, consider two such atoms close together. Suppose that at a certain instant atom 1 has a nonzero dipole moment. This produces an electric field, which induces a dipole moment in atom 2. This induced dipole is always aligned such that it is attracted by atom 1. (This is illustrated, for one possible orientation of the atoms, in Figure 13.4. For other orientations, see Problem 13.5.) Therefore, as the two dipoles fluctuate, each induces in the other a dipole that it attracts. This is the origin of the van der Waals attraction between any two atoms or molecules. [FIG 13.4 = OLD 17.2 XX] * The terminology here is a little confused. A few authors use "van der Waals" to describe both the attraction between two permanent dipoles and the weaker attraction involving induced dipoles. We follow the majority and reserve "van der Waals" to describe only the weaker induced dipole attraction. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.7 13.3 Crystals and Noncrystals All of the mechanisms that bond atoms (or molecules) together into solids have the same general behavior: At large separations, r , the force between two bonding atoms is attractive, at a certain separation, r R , it is zero, and for r less than R it is repulsive. This behavior is illustrated in Fig. 13.5, which shows the potential energy of two atoms as a function of their separation r. [FIG 13.5 = old 17.3 XX] When just two atoms interact as in Fig. 13.5, they have minimum possible energy, and hence form a stable bound state, at separation R. Three such atoms would be in equilibrium at the corners of an equilateral triangle, with the distance between each pair equal to R. Similarly, four would be in equilibrium at the corners of a regular tetrahedron. However, with five or more atoms, it is impossible to arrange them so that the distances between all pairs are the same. Instead, the atoms must arrange themselves so that any one atom is closest to a few others and farther away from all the rest. It is one of the tasks of solid-state physics to predict, or at least explain, the arrangement of the many atoms in a solid that minimizes their total energy and hence makes the solid stable. As one might expect, the stable configuration of many solids has the atoms arranged in a regular pattern, in which the grouping of atoms is repeated over and over again. This regular pattern is called a lattice, and the resulting solid is called a crystal. Experimentally, the structure of crystals is found using diffraction of X rays, neutrons, and electrons, and by the use of various kinds of microscopes (Figure 13.6). The mathematical study of the possible crystal structures is an important part of theoretical solid-state physics. Here we shall just describe a few such structures. [FIG 13.6 = old 17.4 XX] The simplest crystal structure is the cubic lattice, in which the atoms are arranged at the corners of many adjacent cubes, as shown in Fig. 13.7. In a perfect cubic lattice, all these cubes are identical, and each is placed in the same relation to its neighbors. Therefore, all the information about the structure of the lattice is contained in the single cube, or unit cell, shown in Figure 13.7(c). [FIG 13.7 = old 17.5XX] A slightly more complicated crystal lattice is the face-centered cubic (or FCC) lattice, with atoms on each face of a cube as well as at the corners, as in Figure 13.8(a). This arrangement actually gives a slightly higher density of atoms (more atoms per unit volume) than the simple cubic, and is therefore favored in many solids. For example, common salt (NaCl) is an FCC crystal. As shown in Figure 13.8(b), the Cl– ions are arranged on one FCC lattice. The Na+ ions lie on a second FCC lattice, which is offset from the Cl– lattice. In this way, each Na+ is nearest to six Cl– ions, and vice versa. [FIG 13.8 = old 17.5 XX] 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.8 EXAMPLE 13.1 Given that the nearest-neighbor separation in NaCl is r0 0.28 nm , estimate the atomic cohesive energy of NaCl in eV per atom. Let us consider first the potential energy of any one ion, for instance, the Na+ ion at the center of the cube in Figure 13.8(b). This ion has six nearest-neighbor Cl– ions, which give it a negative potential energy of 6ke 2 / r0 . However, the Na+ ion also has 12 next-nearest-neighbor Na+ ions at a distance of r0 2 . (These are all of the remaining 12 Na+ ions in the picture.) These contribute a positive potential energy of 12ke 2 / (r0 2) . Next, there are eight Cl– ions (at the eight corners of the cube in the picture) at a distance of r0 3 , and these contribute a negative potential energy of 8ke 2 / (r0 3) . Continuing in this way we would find a total potential energy: ke 2 12 8 ke 2 t ot al P E of any one ion 6 ... . (13.1) r0 2 3 r0 The sum of the infinite series, , is called the Madelung constant. Since the series is difficult to evaluate, we shall simply state the result, that its sum is 1.75 (to three significant figures). Therefore, the potential energy of any one ion in solid NaCl is ke 2 1.44 eV nm t ot al P E of any one ion 1. 75 r0 0.28 nm (13.2) 9.0 eV. In this calculation we have ignored the short-range repulsive forces due to the overlap of the electron distributions when the ions get very close. The repulsive contribution to the energy is actually rather small (for exactly the same reason as it was in ionic molecules; see Section 13.2). Therefore, our estimate (13.2) is just a little below the observed value t ot al P E of any one ion 7. 9 eV. (13.3) For the remainder of this calculation, we shall use this observed value. To find the energy of the whole NaCl crystal, we cannot simply multiply the answer (13.3) by the total number (N) of ions, since this would count each interaction twice over.* Instead, we must multiply by N/2, to give a total potential energy of (-7.9 eV)N/2. Thus the energy to pull the solid NaCl apart into separated Na+ and Cl– ions is (7.9 eV)N/2. Since N/2 is the number of Na+ - Cl– pairs, we can say that energy t o separat e NaCl int o Na + and Cl - ions 7.9 eV per Na + - Cl - pair If we wish to have separate neutral atoms, we must next remove an electron from each C1– ion (which requires 3.6 eV), and then attach it to the Na+ ion (which gives us back 5.1eV). Thus the total work done is * For each pair of ions, i and j, there is a single potential energy of interaction that went into (13.3). However, if we simply multiply (13.3) by N, we will have counted the potential energy of i due to j and of j due to i. In other words, we will have counted every interaction twice. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.9 energy t o separat e NaCl int o Na and Cl at oms ( 7. 9 3. 6 5. 1) eV per pair 6.4 eV per Na-Cl pair. Dividing by 2, we find finally that atomic cohesive energy of NaCl 3. 2 eV/ atom. [FIG 13.9 = old 17.7 XX] Many other solids have the FCC lattice, for example, diamond* and several metals, including copper, silver, and gold. Two other important structures are the body-centered cubic (BCC) and the hexagonal close packed (HCP), both sketched in Figure 13.9. Many metals, such as sodium, potassium, and iron, are BCC; examples of HCP are the solid forms of hydrogen and helium, and several metals, such as magnesium and zinc. MACROSCOPIC APPEARANCE OF CRYSTALS When a large enough number of atoms form into a single crystalline lattice, the resulting solid is a macroscopic crystal. As the crystal forms, its macroscopic surfaces tend to coincide with the planes of its microscopic lattice; if a grown crystal is struck, it tends to cleave along these same atomic planes. Either way, the crystal acquires smooth flat surfaces that reflect light like a mirror. If the material is transparent, these same surfaces can cause strong internal reflections, which add to the glittering appearance that we associate with crystals. There are many examples of such macroscopic crystals: ordinary table salt, gem stones, the fluorite crystal in Figure 13.10, and many more. [FIG 13.10 = old 17.8 XX] Many solids, including most metals, have a microscopic crystalline structure but do not, nevertheless, appear crystalline. This is because, instead of forming as a single large crystal, these solids form in a jumbled array of many tiny microcrystals, called grains, as illustrated in the micrograph of stainless steel in Figure 13.11. Since a typical microcrystal, or grain, is about 1 m across, the crystalline nature of such solids is completely invisible to the naked eye. Nevertheless, each grain contains some 1010 atoms, and, from a microscopic point of view, the dominant feature of the material is its crystalline structure. [FIG 13.11 = old 17.9 XX] * This claim does not contradict our earlier statement that diamond has a tetrahedral structure. The carbon atoms are placed on two FCC lattices, one offset from the other. Each atom on either lattice is covalently bonded to its four nearest-neighbor atoms, all of which are on the other lattice at the corners of a regular tetrahedron. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.10 EXAMPLE 13.2 An important task for solid-state physics is to predict the strengths of materials. Given the estimate in Problem 13.22 for the force needed to separate two bonded atoms, make a rough estimate of the tensile strength of a simple cubic material with lattice spacing r0 0.3 nm . The tensile strength of a material is the force per unit area needed to rupture it. From Problem 13.22XX we find that the force needed to separate a single pair of bonded atoms is of order F 3 109 newt on . The density of atoms in any one crystal plane is 1/ r0 . 2 Therefore (if we ignore all but nearest-neighbor interactions), the force to tear apart a unit 2 area of two adjacent planes is F / r0 ; that is F 3 109 N t ensile st rengt h 10 3 1010 N/ m 2 . (13.4) (3 10 m) 2 2 r0 This calculation assumed a cubic lattice and included only nearest-neighbor interactions. Furthermore, although the force that we used is realistic for strong bonds (covalent, ionic, or metallic), it would be a gross overestimate for weaker bonds such as the van der Waals bond. Nevertheless, the answer (13.4) should — and does — give the right order of magnitude for any single crystal of strongly bonded atoms. Single-crystal whiskers of iron (as in Figure 13.12) and of quartz have tensile strengths of this order. Macroscopic samples of many crystalline materials have tensile strengths much smaller than (13.4) (that for cast iron is some 100 times smaller), but this is because such macroscopic samples are not perfect crystals. It is the many imperfections in these materials that break easily and explain their much lower tensile strengths. [FIG 13.12 = old 17.10 XX] AMORPHOUS SOLIDS Not all solids have a crystalline structure. Instead, the atoms of many solids, such as glass, wax, and rubber, arrange themselves in an irregular pattern. Such solids are described as amorphous, from a Greek word meaning "without form." In fact, many materials can solidify in either a crystalline or an amorphous form, depending on their preparation. For example, a material may form in a crystalline state if cooled slowly from a liquid, a process called annealing. (Slow cooling gives the atoms time to arrange themselves in the lowest-energy, crystalline, state.) On the other hand, the same material may form in an amorphous state if quenched, that is, cooled very rapidly. The difference between crystalline and amorphous solids is shown schematically in Figure 13.13. Part (a) illustrates a perfect crystal, with its atoms in a rigid and regular lattice. Part (b) shows an amorphous solid, with its atoms bonded in a more-or-less rigid structure, but not in a perfectly regular pattern. For comparison, parts (c) and (d) illustrate a liquid and a gas. When one heats a solid sufficiently, it first melts [part (c)]: Although its atoms remain within the range of interatomic forces, they are no longer held in a rigid pattern and are, instead, free to move around. When heated still further, the liquid 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.11 becomes a gas [part (d)]: The atoms separate much further, they move more rapidly, and they interact only during their brief collisions. [Fig.13.13 = old 17.11 XX] We can characterize the differences between crystals, amorphous solids, liquids, and gases in terms of their long-range order. A perfect crystal has perfect long-range order: If we know its crystal structure and the location of one atom, we can predict with certainty the location of all other atoms. This is illustrated in Figure 13.14(a), which shows the probability density (the probability of finding an atom) plotted along one crystal axis for an ideal crystal. If we know that there is an atom at the origin, we can predict with certainty that there are atoms at r0 , 2 r0 , 3 r0 , . . . , for as far as the solid extends.* [Fig.13.14 = old 17.12 XX] The distribution function for an amorphous solid is like Figure 13.14(b). If we know that there is an atom at the origin, the probability of finding another atom very close to r 0 is small, since two atoms repel one another strongly at short range. The probability increases to a peak at about the same r0 at which the next atom of a crystal would be located. It then drops again because the repulsion between overlapping atoms prevents another atom from locating too close to the atom of the first peak. Because of the amorphous solid's irregular structure, we can predict the atoms' positions with less and less certainty as we move on, and the density function rapidly approaches a constant, reflecting our complete ignorance of the positions of the more distant atoms. We say that the amorphous solid shows short-range order, but no long-range order. The probability density for a liquid is almost exactly like that for an amorphous solid, shown in Figure 13.14(b). In fact, an amorphous solid can be regarded as liquid, "frozen in time." The probability density for a gas is shown in part (c). It is much lower than those for solids or liquids, since the density is much less; it shows even less structure than that for the amorphous solid because a gas has even less order. 13.4 Energy Levels of Electrons in a Solid; Bands We have stated that the reason metals conduct electricity is that some of their conduction electrons can move essentially freely within the boundaries of the metal, but this does not explain why the electrons in a metal can move so freely, nor why the electrons in an insulator cannot. To explain this difference we must examine the energy levels for the electrons in a solid. Fortunately, we can do this quite easily, leaning on our discussion of the energy levels of electrons in molecules (Section 12.4). As one might expect, once we * Strictly speaking, Fig. 13.14(a) represents a classical crystal at absolute zero. In quantum mechanics the uncertainty principle means that we cannot know the locations with complete certainty, and at nonzero temperatures thermal motions introduce a further uncertainty. For both reasons the ideal spikes of Fig. 13.14 (a) should be a little spread out. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.12 understand the energy levels of the electrons, we can explain many properties of solids in addition to their conductivity, In Section 12.4 we described the energy levels of an electron belonging to either of two atoms as the atoms come together to form a molecule. While the atoms are far apart, we can focus on any one atomic orbital (1s, 2s, . . . ). Since the electron can occupy this orbital in either atom, this gives two degenerate orbitals (that is, two possible wave functions with the same energy). We saw in Section 12.4 that as the two atoms come closer together and begin to overlap, this degeneracy splits, and we have instead two orbitals of different energy, as indicated in Figure 13.15(a). [FIG 13.15 = old 17.13 XX] If we start instead with three atoms far apart, then for each atomic orbital there are three independent, degenerate wave functions for an electron (corresponding to attaching the electron to any of the three atoms). As we move the atoms together this threefold degeneracy is split, and we get three distinct possible energies, as indicated in Figure 13.15(b). More generally, if we consider N atoms far apart, then for each atomic orbital there are N degenerate wave functions, and as we move the atoms together these fan out into N more-or-less evenly spaced levels. We have shown this in Figure 13.15(c) where we have indicated the many closely spaced levels by shading. We can imagine putting together a whole solid in this way, starting with N well-separated atoms. While they are still far apart, we can arrange the atoms in the same geometrical pattern as the final solid, and we can then imagine the whole structure slowly contracting to its equilibrium size. As the nearest-neighbor separation r decreases, the energy levels of the electrons will vary, their general behavior being that shown in Figure 13.15(c), which we have redrawn in more detail as Figure 13.16. For each atomic orbital there are N independent wave functions, all of which have the same energy when r is large. As we reduce r, this degenerate level fans out into N distinct levels. We refer to these N closely spaced levels, which all evolved from a single atomic level, as a band. The width and spacing of these bands varies with r, but the bands of the actual solid are found by setting r equal to its equilibrium value r0 , as shown on the left of Figure 13.16. [FIG 13.16 = old 17.14 XX] Figure 13.16 introduces some of the terminology used in connection with the energy levels of electrons in solids. The band of levels that evolved from the 1s atomic levels is called the 1s band, and similarly with 2s, 2p, and so on. The range of energies in a single band — the width of the band — is typically a few eV. If there are N levels within the band,* the average spacing E between adjacent levels is a few eV E . N * As we shall see shortly, there are generally more than N states in a band, but this only strengthens our conclusion that the average spacing is very small. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.13 In practice, N is very large; even in a microcrystal 1 m across, there are some 1010 atoms (Problem 13.25XX). Thus the level spacing within a band is very small; in a microcrystal with N 1010 , E is of order 10–10 eV, and in a larger crystal it is even smaller. Therefore, for almost all practical purposes, we can think of the electron energy levels in a solid as continuously distributed within each band. This is what the shading of the bands in Figure 13.16 was intended to suggest. It can happen that the highest level in one band is above the lowest level in the next-higher band. In this case we say that the two bands overlap. On the other hand, it often happens (as in Fig. 13.16) that two neighboring bands do not overlap, and there is instead a band gap — a range of energies that are not allowed. We shall see that the occurrence of band gaps, and their widths, are two of the main factors that determine the conductive properties of a solid. It is important to know whether all the states in a given band of a solid are occupied (just as it was important in our discussion of the periodic table to know whether all the states of a given level in an atom were occupied). To this end, we need to know how many states there are in a given band. For s bands (the 1s and 2s bands of Figure 13.16, for example) there are N independent wave functions and two possible spin orientations, and hence 2N states in all. If we consider a p band (a band that evolved from an l 1 atomic level), then when the N atoms were well separated there were 3N independent wave functions (three different values of lz for each atom) and two spin orientations. Therefore, there are altogether 6N states in a p band. Since we shall only be concerned with s and p bands, we shall not give a general formula here (but see Problem 13.28XX). 13.5 Conductors and insulators – a qualitative view We are now ready to explain the difference between conductors and insulators. We illustrate our discussion with three examples, starting with lithium, which we know to be a metal and hence a conductor. LITHIUM: A CONDUCTOR The lithium atom has the configuration 1s22s1, with its 1s level full and its 2s valence level half full. As we assemble N lithium atoms into solid Li, the 1s and 2s levels fan out into the 1s and 2s bands, which we have shown in Figure 13.17. The N lithium atoms contribute altogether 3N electrons, while each of the two bands can hold 2N electrons. Thus the 1s band is completely filled, and the 2s valence band is half filled. We have shown this occupancy in Figure 13.17, where dark shading represents occupied, and light shading unoccupied levels. The energy shown as E F is the Fermi energy, which is defined as the energy of the highest occupied level.* * Strictly speaking, this is correct only at absolute zero. At higher temperatures, thermal motions can excite a few electrons to levels somewhat above EF as describe in Section 13.10. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.14 [FIG 13.17 = old 17.15 XX] We are now going to argue that the electrons in the partially full valence band can move around in response to an applied electric field — they are conduction electrons, so the partially full valence band is also called the conduction band — whereas the electrons in the full 1s band cannot move. To see this, let us consider a lithium wire oriented along the x axis. In the absence of an electric field, the energy levels are the same at all points along the wire, as indicated in Figure 13.18(a). [FIG 13.18 = old 17.16 XX] Suppose now that we connect a battery to the ends of the wire to maintain an electric field to the right. This field raises the potential energy of the electrons on the right, and the bands now look like Figure 13.18(b) or, in a magnified version, Figure 13.18(c). For any electron at the top of the occupied levels in the conduction band, there are now unoccupied levels of the same energy (and lower) just to the left. Therefore, each of these electrons begins to move to the left in response to the electric field, as indicated (for one electron) by the horizontal arrow in (c). This motion of electrons to the left is, of course, the expected electric current. In due course, the electrons moving to the left collide with defects and other "imperfections" of the lattice and lose some kinetic energy, as indicated by the downward arrow in Figure 13.18(c). These collisions, which are discussed in Section 13.8, are the main cause of electrical resistance in most metals. The sequence of accelerations and collisions is continually repeated as the electrons move along the wire. The energy lost by the electrons in collisions increases the vibrations of the lattice, and the wire's temperature rises. This temperature rise is just the familiar Joule heating of a wire by a current. The argument just given does not apply to electrons in the full 1s band, since there are no vacant states to their immediate left. These electrons are immobilized by the complete lack of empty states to transfer into. Thus the electrons in a filled band do not contribute to conduction. This example already illustrates our two most important conclusions: Solids with a partially filled band are conductors; those in which all occupied bands are 100% full are insulators. Our next two examples show that these simple rules have to be applied with care. BERYLLIUM: A CONDUCTOR We turn now from lithium to the next element in the periodic table, beryllium. Its atomic configuration is 1s22s2, and both of its occupied levels are filled. This might suggest that solid Be should have filled 1s and 2s bands and should therefore be an insulator, whereas Be is, in fact, a good conductor. The explanation of this apparent contradiction is not hard to find: As we assemble solid Be and its atomic levels fan out into bands, the 2s and 2p bands spread so far that they overlap one another, as shown in Figure 13.19. The 2s and 2p bands thus form a single valence band, which can accommodate altogether 8N electrons. (Remember an s band holds 2N electrons and a p 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.15 band 6N.) Thus, of the 4N electrons in the solid, 2N fill the 1s band, and the remaining 2N can only partially fill the valence band. Therefore, Be is a conductor. [FIG 13.19 = old 17.17 XX] DIAMOND: AN INSULATOR Atomic carbon has the configuration 1s22s22p2, with its n 2 , valence shell half filled. This suggests that solid carbon should be a conductor, which, indeed, it is in the form of graphite. However, solid carbon in the form of diamond is an outstanding insulator. To explain this puzzle, we must look carefully at the behavior of the energy levels as we assemble a diamond. In diamond the carbon atoms are covalently bonded. As we bring the atoms together and the covalent bonds form, the 2s and 2p levels fan out into bands, as we would expect. However, each level splits into two separate bands, as shown in Figure 13.20, (This is closely analogous to the development of the bonding and antibonding orbitals described in Section 12.4.) At the equilibrium separation r0 , the lower two bands overlap each other, as do the upper two. Thus the band structure of diamond is as shown on the left of Figure 13.20: There is an n 1 band that can hold 2N electrons, and then two well-separated n 2 bands, each of which can hold 4N electrons. Since carbon has 6N electrons in all, the lowest two bands are completely full, and the upper n 2 band is completely empty. Because the lower n 2 band holds all the valence electrons, it is called the valence band, and because electrons in the upper n 2 band (if there were any) would contribute to conduction, this band is called the conduction band. With its valence band full and its conduction band empty, diamond is an insulator. It turns out that as indicated in Figure 13.20, the gap between the valence and conduction bands of diamond is about 7 eV. Under normal conditions, this means that no electrons can be excited into the empty conduction band. (At room temperature, for example, thermal motions could excite an electron, but the probability is so low that the expected number of excited electrons in a large diamond is far less than 1.) This makes diamond an extremely good insulator. This same large band gap also explains why diamond is transparent. Visible photons have energies of 2 to 3 eV and cannot excite any electrons across the 7 eV gap; therefore, visible light cannot be absorbed in diamond. [FIG 13.20 = old 17.18 XX] We mentioned in Section 13.2 that solid silicon (just below carbon in the periodic table) has the same crystal structure as diamond. The 3s and 3p levels in Si behave almost exactly like the 2s and 2p levels of diamond, and silicon, like diamond, is a nonconductor. There is, however, a subtle but crucial difference. The equilibrium separation r0 in Si is larger than in diamond (0.24 nm in Si compared to 0.15 nm in diamond). It is easy to see in Figure 13.20 that a larger value of r0 suggests a smaller band gap, and, indeed, the band gap in silicon is only about 1 eV (compared with 7 eV in diamond). Since visible photons can easily excite electrons across a 1 eV gap, light is quickly absorbed in silicon, which is therefore not transparent. The narrow band gap in silicon also has a dramatic effect on its conductivity. At low temperatures, all the valence electrons in Si are locked in the full valence band, and Si is an insulator. But as the 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.16 temperature rises, a substantial number of electrons are excited across the narrow gap into the conduction band, and the conductivity increases rapidly. For this reason, silicon and other solids with a similar narrow gap between a full valence band and an empty conduction band are called semiconductors. Semiconductors include silicon and germanium, both from group IV of the periodic table, and several compounds of group III with group V, such as gallium arsinide. Semiconductors and their enormous technological importance are discussed in greater detail in Chapter 14. 13.6 The Drude Model of Conductivity In this and the following two sections, we look more closely at the conductivity of metals. We describe a simple but powerful model of electrical conduction, called the Drude* model. The Drude model is an example of a "semi-classical" model, meaning it is a hybrid of classical and quantum ideas. In the Drude model, we think of conduction electrons as classical particles with a definite position and velocity, but to understand the interaction of these electrons with their surroundings, we use the results of quantum calculations. The classical view of electrons as point particles can be made roughly consistent with the quantum view of electrons as de-localized wavefunctions through the notion of wavepackets, described in section 6.7XX. A wavepacket describing an electron in a metal can have both an average position and an average velocity, and so, within the constraints set by the Heisenburg uncertainty principle, we are allowed to view electrons in a metal as localized particles. See Problem 13.XX. Before describing the Drude model, we begin with some definitions. The electrical conductivity of material is defined by the equation j E . (13.5) Here, E is the electric field† and j is the current density, which is defined as the current I per area A, that is, j = I / A. Both the current density and the electric field are vectors, but, since we will be working in one dimension, we will drop the vector notation and use signs to represent direction, with (+) meaning rightward and (–) meaning leftward. In the SI system, conductivity has the units of (ohmm)-1, although one often sees the units of (ohmcm)-1 used. The resistivity of a material is defined as the inverse of the conductivity, 1/ , so resistivity has units of ohmm or ohmcm. Equation (13.5) can then be rewritten as E j . (13.6) This equation is the microscopic version of Ohm's Law V = IR , which is discussed in Sec13.8. In Problem 13.X, you will show that (13.6) is indeed equivalent to Ohm's Law. * Pronouced ―DREW-dah.‖ † Here we use E for electric field, whereas, in earlier chapters, we used to avoid confusion with the symbol for energy. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.17 A good electrical conductor, like copper, has a large and a small , while a good electrical insulator, like glass, has an extremely small and a huge . Table 13.1 and Figure 13.21 shows the conductivities and resistivities of various materials. Notice that conductivity varies over some 24 orders of magnitude — an enormous range — from 107 or 108(ohmm)–1 for good metals to 10–16 (ohmm)–1 for insulators. [FIG 13.21 New XX] In 1900, just 3 years after J. J. Thompson discovered the electron and well before the development of quantum mechanics, Paul Drude (German, 1863-1906) proposed a model of electrical conduction in metals, a model which is still used today because of its simplicity and usefulness. Drude made the bold assumption that the electrons in a metal behave as a gas of free particles. These free electrons, which we now understand are the conduction electrons, accelerate when an electric field is applied. If there were no frictional forces present, then the electrons would accelerate without limit, leading to an ever-increasing current. Since such runaway currents are never observed, there must be some mechanism at work which acts to slow the electrons. Drude thought that the electrons must collide with the nuclei of atoms, but this turns out to be incorrect. Because of a quantum mechanical effect, which we shall describe later, the electrons do not collide with the nuclei but instead they collide with both "impurity atoms" and "lattice vibrations." These terms will be explained in detail in the next section. Regardless of what the electrons collide with, we can define a scattering time as an average time between collisions. Between collisions, an electron with charge q = e in an electric field E will experience a force F = q E = – e E and hence an acceleration a = F/m = – e E/m, where m is the mass of the electron. If the electron starts from rest, then, in a time t, the electron will acquire a velocity eE v at t. (13.7) m Here, the minus sign indicates that, for negatively charged electrons, the direction of the velocity is opposite to the direction of the electric field. The effect of collisions, whatever their origin, is to momentarily interrupt this free acceleration and decrease, stop, or reverse the forward velocity. Our picture of the electron motion is then one of a very jerky, start-stop kind of motion. In the competition between the speeding up due to the E- field and the slowing down due to collisions, the electron acquires an average velocity called the drift velocity, vd . With equation (13.7) in mind, we now define the scattering time by the relation eE vd . (13.8) m The scattering time is the average time between collisions. A given electron actually experiences a wide range of time intervals between successive collisions, and properly defining an "average" time interval has some surprising subtleties which are explored in Problem 13.xx. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.18 Drude related this scattering time to the electrical conductivity by the following argument. First, we show that the current density j can be written as j n q vd . (13.9) where n is the concentration (number per volume) of charge carriers, q is the charge per carrier, and vd is the drift velocity. To understand this equation, consider a wire carrying a current I, as in Figure13.22. In a time t , all the charges will move a distance x vd t as shown. A cylindrical portion of the wire of length x and cross-sectional area A contains a total charge [FIG 13.22 new XX] Q = (charge per carrier) (number of carriers) = (charge per carrier) (number per volume) (volume) = q n ( x A ) q n (vd t A ). All the charges making up this total charge Q will pass through the right end of the cylinder during a time interval t , and the resulting current is I Q / t . The current density is then I Q j q n vd , A t A and we have demonstrated (13.9). Now inserting (13.8) into (13.9), and setting q = – e, we obtain eE ne 2 j nq E . (13.10) m m Comparing (13.10) with the equation j = E, we conclude that ne 2 (13.11) m This equation is the central result of Drude's model of conductivity. It says that the conductivity is large when there are many charge carriers per volume (large n), few collisions (large ), and light charge carriers (small m). The dependence on mass can be understood intuitively by noting that low-mass carriers acclerate more quickly in an applied field ( a qE / m ), leading to a larger . Consequently, ionic conductors such as electrolytic solutions, in which massive ions carry the current, have a much smaller conductivity than normal metals, in which light electrons are the charge carriers, as seen in Table 13.2. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.19 Example13.3. A copper wire of diameter 1.0 mm (radius r = 0.5 mm) carries a current of I = 10 A. Estimate the magnitude of the drift velocity vd. According to (13.9), j n q v d , the drift velocity is easily computed if the current density j and the carrier concentration n are known. The current density is readily calculated as I I 10 A j 1.27 105 A/ m 2 12. 7 A/ cm 2 . A r 2 (5 103 m) 2 (Here, we give the number in both SI units, as well in the more practical units of A/cm2, often used by electrical engineers.) Computing n, the number density of conduction electrons in copper, is not quite so easy. One can compute it roughly by noting that, for most solids, the nearest-neighbor atoms are about r0 = 0.3 nm apart and for copper, the valence or number of conduction electrons per atom is 1, as one might surmise by noting its electronic configuration — a single s electron outside a d shell. We conclude that the number density of charge carriers for copper is about 1 1 n 4 1028 m -3 , r03 ( 0.3 109 m) 3 (The measured value is actually a factor of 2 higher than this.) Finally, we can compute the drift velocity as j 1.27 105 A/ m 2 vd 3 19 2 105 m/ s 0.02 mm/ s . nq ( 4 10 m )(1.6 10 C) 28 This is a very tiny speed; a given electron in this wire takes more than a day to travel a meter! When you flip a light switch, the electromagnetic signal travels along the wire at nearly the speed of light, but the electrons themselves move with glacial slowness. A long garden hose filled with water behaves similarly. When you open the faucet, water comes out the far end of the hose immediately, due to the incompressibility of water, while a given water molecule takes several seconds to travel the length of the hose. 13.7 Electron Collisions in Metals We now return to the nature of the collisions experienced by conduction electrons in a metal. Understanding the origin of these collisions requires a knowledge of quantum mechanics, which did not exist in Drude's time. In a metal, the atomic nuclei surrounded by their inner core electrons form an array of positively charged ion cores. A conduction electron sees these ion cores as an array of attractive potential wells separated by barriers, as shown in Figure 13.23. To understand how an electron interacts with these barriers, we must recall some qualitative features of quantum mechanical tunneling, which was discussed in Chapter 7. [FIG 13.23 new XX] When a free electron wave packet encounters a single potential barrier, as in Figure13.24(a), there is, in general, a reflected wave and a transmitted wave. If the barrier 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.20 height U is high enough or the barrier width L is sufficiently large, the electron will be almost entirely reflected. If there are two barriers, as in Figure13.23(b), then a remarkable phenomena called resonant tunneling can occur. If the energy of the incident electron wave has certain special values, then the waves reflected from the two barriers will interfere destructively, resulting in no net reflected wave and 100% transmission of the incident wave. There is an analogous process in the field of optics: an anti-reflection coating on a glass surface. If the coating has the correct thickness and refractive index, then an incoming wave of a particular frequency experiences 100% transmission and no reflection. [FIG 13.24 new XX] If instead of two barriers, there is an infinite array of uniformly spaced identical barriers, as in Figure 13.23, then it turns out that resonant transmission occurs over a range of electron energies. In this band of energies, the electron behaves essentially as a free electron. It can travel forever, without scattering. This range of energies in which free propagation occurs is precisely what we called the conduction band in Section 13.5. If the nuclei in a metal formed a perfectly regular array, then, because of resonant tunneling, the conduction electrons could zip right through the metal unimpeded, and the metal would have infinite conductivity. However, any disruption in the periodicity of the lattice causes electron scattering, and there are two causes of non-periodicity which are always present in a metal. First, any metal sample always contains some impurities, no matter how carefully it has been purified. Impurities may be foreign atoms, or they may be defects, such as vacancies (missing atoms), interstitial atoms, or grain boundaries, all of which are illustrated in Figure 13.25. [FIG 13.25 new XX] Besides impurities and other defects, lattice vibrations also disrupt the periodicity of the lattice and cause electron scattering. At any nonzero temperature, the atoms of a solid vibrate about their equilibrium positions, as in the ball-and-spring model of a solid shown in Figure 13.26. The higher the temperature, the more vigorous the vibrations. At a high enough temperature, this "jiggling" of the atoms becomes so great that the solid melts. In the correct quantum mechanical description of this ball-and-spring model of a solid, this vibrational energy is quantized and the quanta of energy are called "phonons". In classical language, higher temperature means larger amplitude vibrations, while in the language of quantum mechanics, higher temperature means more phonons, and consequently more electron-phonon scattering. It turns out that at room temperature and above (T 300K), the rate of electron-phonon scattering is proportional to temperature, leading to a resistivity proportional to temperature. [FIG 13.26 new XX] In Figure13.27, we indicate schematically the potential that a conduction electron encounters in three environments: (a) a perfectly periodic potential such as is almost acheived in very pure metals at very low temperatures; (b) the non-periodic potential seen in a metal with some impurities, in which case different types of atoms produce different well-depths; and (c) the non-periodic potential found in a pure sample at high 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.21 temperatures, in which case lattice vibrations produce a non-uniform spacing of the atoms. In cases (b) and (c), compared to case (a), there is much more scattering of electrons, a shorter time between collisions, and a smaller conductivity = ne2 /m. [FIG 13.27 new XX] [FIG 13.28 new XX] The general shape of the curves of resistivity vs. temperature found in metal samples are displayed Figure13.28. Three curves are shown corresponding to three different samples: (a) A "clean" sample, such as moderately pure copper or aluminum. In this case, at higher temperatures (room temperature and above), lattice vibrations are the dominant source of electron scattering leading to a linear dependence of on T. At very low temperatures (a few ten's of Kelvin or lower) lattice vibrations are so weak that impurities dominate electron scattering, leading to a temperature-independent resistivity. (b) An "ordinary" sample, very like (a) but not so pure. Here, the impurity scattering is larger than in case (a) leading to a higher resistivity at low T. The dependence of the low- temperature resistivity on purity leads to a simple way to quantify the purity of samples: the residual resistivity ratio (RRR) or "triple-R" is defined as the ratio of the resistivity at room temperature to that a helium temperatures (a few Kelvin). The higher the triple-R, the more pure the sample. Carefully purified metal samples have a triple-R of a few hundred; very dirty samples have a triple-R of less than 2. (c) A "dirty" (very impure) metal or an alloy. An alloy is a mixture of metals; an example is brass, which is a mixture of copper and tin. In an alloy, almost every atom can be regarded as an "impurity" and consequently the impurity scattering is so large that it dominates the resistivity at all but the highest temperatures. Another example in this category is stainless steel, which is iron with a very high concentration of other elements including chromium and nickel. Stainless steel can be regarded as an alloy or as very "dirty" iron . In Figure 14.29, we plot the resistivity against temperature T for a typical metal such as copper and for a pure semiconductor, such as silicon. Drude's equation ne 2 / m gives us insight into the very different behaviors of metals and semiconductors seen in this graph. In metals, the temperature dependence of the conductivity is due entirely to the temperature dependence of the scattering time . The carrier concentration n is temperature-independent, while the scattering time decreases with increasing temperature. As the temperature increases, decreases due to increased scattering from lattice vibrations; consequently, the conductivity decreases and the resistivity rises,. In semiconductors, however, the temperature dependence of the resistivity is due mostly to the temperature dependence of the carrier concentration. In semiconductors, n is small compared to metals, and very strongly temperature-dependent. The smallness of n results in a large resistivity compared to metals. In pure (undoped) semiconductors, the carrier concentration n is thermally activated: as the temperature is increased, more and more electrons are thermally excited from the valence band into the 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.22 conduction band, across the band gap; n increases rapidly with increasing temperature, leading to a decrease in resistivity. [FIG 13.29 new XX] 13.8 The Fermi Speed A very important feature of the conductivity of metals, a feature we have not yet discussed, is that the conductivity in the equation j E is independent of the applied electric field E or the current density j. That is to say, although j and E may vary considerably, the ratio j/E remains constant. Conductors which exhibit a field- independent conductivity are called ohmic conductors. Resistors made from ohmic conductors obey Ohm's Law, which is the statement that V/I = R = constant, where I is the current through the sample, V is voltage across it, and R is its resistance. It is important to note that the equation R = V/I , by itself, is not Ohm's Law; it is merely the definition of the resistance R of a sample. Ohm's Law is the statement that the resistance (defined as V/I) is a constant, independent of V or I. Most conductors are ohmic, so long as the current density j is low enough that Joule heating does not raise the temperature of the sample. If significant Joule heating does occur (as in most incandescent light bulbs), then the increased temperature causes an increase in due to increased electron-phonon scattering, and the sample under study exhibits non-ohmic behavior. The fact that the conductivity of a metal is independent of the applied electric field E is surprising for the following reason. The conductivity depends on the scattering time according to the Drude formula (13.11), ne 2 / m . At first glance, it might seem that the scattering time should depend on the field E. After all, if the field E is increased, the acceleration a of a conduction electron will increase (according to a = qE/m), and if the electron accelerates more quickly it will surely encounter an impurity or phonon more quickly, resulting in a shorter collision time and a smaller conductivity . This argument, which implies that depends on E, would be correct, if conduction electrons had zero initial velocity after a collision. However, due to a quantum mechanical effect which we explain shortly, the electrons in a metal always have a very large intrinsic speed, called the Fermi speed vF. The Fermi speed is about 0.3% of the speed of light ( vF 0.003c 106 m/ s ). In contrast, the magnitude of the drift velocity vd , which is the mean velocity due to the applied field E, is miniscule ( vd 0.01 mm/ s , as in Example 13.1). Consequently, the increase in velocity due to the E field (and hence the decrease in time until the next collision) is completely negligible compared to the initial huge Fermi speed of the conduction electron, and the scattering time is almost completely independent of the field E. This lack of dependence of on E is the origin of Ohm's Law. The motion of conduction electrons is thus one of extremely rapid, random, motion due to the large Fermi speed and, superimposed on this, a very small bias toward one direction due to the tiny drift velocity. As shown in Figure 13.30, a conduction electron in a metal follows a random zig-zag path even with zero applied field (E = 0), regardless of the temperature. This is an example of random walk diffusive motion , 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.23 which was described in Sec.3.8. The mean free path* l is defined as the mean distance an electron travels between collisions, that is, the step length in the random walk. The mean free path is related to the Fermi speed and the scattering time by l vF . (13.12) The mean free path (mfp) depends on the purity and temperature of the sample. Very dirty samples, such as alloys, or metals at very high temperatures, may have a mfp of nanometers or less. Metals of ordinary purity at room temperature have a mfp of roughly 10-100 nm. While high-purity (―clean‖) samples at low temperatures may have a mfp of many micrometers or even millimeters. [FIG 13.30 new XX] Example 13.4. Using the Drude model and the known Fermi speed, estimate the scattering time and the mean free path for conduction electrons in copper at room temperature. Drude's equation, ne 2 / m , allows us to compute the scattering time, , from the measured conductivity, , given in Table 13.2. From our previous Example (13.3), the conduction electron density is about n = 8 1028 m–3. Thus, m (9.11 1031 kg)[5.9 107 ( m) 1 ] 3 19 3 1014 s . n e2 (8 10 m )(1.6 10 C) 28 2 Assuming that the Fermi speed is approximately 106 m/s, we have for the mfp l vF (106 m/ s)(3 1014 s) 3 108 m 30 nm . This is approximately 100 lattice constants. (A lattice constant is defined as the linear size of a ―unit cell‖ in a crystal; it is roughly equal to the distance between nearest- neighbor atoms.) We now consider why conduction electrons in metals have such a large intrinsic speed, the Fermi speed. We first note that the Fermi speed is not simply the thermal speed of the electrons. As described in Section 3.7, in a classical gas of particles of mass m at temperature T, the average thermal speed v of the particles is given by the expression 1 mv 2 2 kT . For electrons at room temperature, this thermal speed is about 2 3 105 m/s, a factor of 10 smaller than the Fermi speed. Another difference between the thermal and Fermi speeds is that the Fermi speed is independent of temperature, while the thermal speed rises with increasing temperature. The large Fermi speed of conduction electrons is a consequence of the Pauli exclusion principle. To see why, we begin by noting that conduction electrons in a metal * In Section 3.8, we used the symbol for the mean free path. Here, we use the symbol l, to avoid confusion with the de Broglie wavelength. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.24 can be regarded as free electrons in a box, a problem which was considered in Chapters 7 and 8. Of course, the electrons are not truly free, that is, they are not in a region where the potential energy function U(x) is constant. An individual electron experiences the attraction of the ion cores of the atoms in the sample, as well as a repulsive potential energy due to the other conduction electrons. However, the Coulomb repulsion from the other electrons nearly cancels the Coulomb attraction to the ion cores, and the electrons behave as a "nearly free" particles. Let us imagine placing electrons in an empty box, one at a time. Fig.13.31 shows the wave functions of the various electron states for the case of a one-dimensional box of length L. The first two electrons will occupy the ground state (n = 1) with one electron spin up and the other spin down. Because of the Pauli exclusion principle, the next two electrons will be forced to occupy the first excited state (n = 2); the next two will occupy the 2nd excited state (n = 3), etc. As more electrons are poured into the box, they are forced to occupy higher and higher energy states. This filling of states is the partial filling of conduction band states described in Section 13.5. If there are very many electrons in the box (many electrons in the conduction band) then the average energy of an electron will be quite large compared to thermal energies. This is the reason that conduction electrons in a metal are so energetic. In a metal, the density of conduction electrons is such that all of the quantum states up to a high energy level are occupied. As we said in Section 13.5, this highest occupied level is the Fermi level, and its energy is called the 2 Fermi energy EF = 1 mv F . (As you might guess, all this was first explained by the Italian- 2 American physicist Enrico Fermi.) In a metal, the Fermi energy has a value of a few electron volts, and, this corresponds to a speed, the Fermi speed, of about 106 m/s. [FIG 13.31 new XX] We can estimate the value of the Fermi energy by the following argument. Notice, in Figure13.31, that the nth state of the system has a wavelength which satisfies n L. (13.13) 2 When there are N electrons in the system, the highest occupied state is the state with n = N/2, so we can write N L. (13.14) 4 If we temporarily regard electrons as classical (rather than quantum) objects, then the average distance between nearest neighbor electrons is r = L/N, which by eqn.(13.14), becomes r = / 4 . Although we have made this argument in one dimension (1D), a similar result is true in 3D; that is, in either 1D or 3D, the average distance between electrons is very roughly equal to the wavelength of the highest occupied state. In 3D, a careful treatment gives r / 2 , rather than the 1D result, r / 4 . We can now compute the kinetic energy of electrons at the Fermi energy. We write 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.25 p2 h2 h2 h2 E (13.15) 2 m 2 2 2m 2m 2r 8 m r2 where we have taken r / 2 as the nearest neighbor distance. If we consider a monovalent metal, that is, a metal with one conduction electron per atom, then the average distance r between conduction electrons is the same as the average distance between atoms, which is about 0.3 nm for most solids. We then have for the Fermi energy, h2 ( 6.6 1034 J s)2 EF 8 9.1 1031 kg 0.3 109 m 2 8 m r2 (13.16) 19 7 10 J 4 eV The Fermi energy varies from metal to metal due to variations in the valence of the metal and the density of the atoms, but it always falls in the range from 2 to 12 eV. This is the same order of magnitude as the energy of chemical bonds, so the result is not too surprising. We can now, finally, compute the Fermi speed. Since E F 1 m vF2, 2 vF 2E F / m 2 7 1019 J / 9.1 1031 kg 1.2 106 m/ s. (13.17). Note, once again, how enormous this speed is compared to the drift speed. We conclude our discussion of the Drude model with the story of a curious historical accident. As we have mentioned before, Drude's work was done well before the advent of quantum mechanics, so there was no way for Drude to compute the Fermi speed or the mean free path. Why then was Drude's model immediately successful? It was because Drude's calculations contained two large mistakes which cancelled! From the known values of the conductivity of metals, Drude used his formula (13.11) to correctly compute the scattering time . As a check on the theory, he also computed the scattering time using = l/v, where l is the mean free path and v is the mean electron speed. But his values for l and v where both too small by an order of magnitude. He assumed, incorrectly, that the electrons scattered from atomic nuclei, so his estimate of the mean free path l was about a nanometer, smaller by a factor of at least 10 than the correct value for metals at room temperature. He also incorrectly assumed that the speed of the electrons is given by the thermal speed of about 105 m/s, which is about 10 times smaller than the correct Fermi speed. Because both numbers were wrong by about the same factor, the ratio, , came out correct. Consequently, the model appeared to be nicely self-consistent and it was immediately accepted by physicists. More than 30 years passed before enough quantum mechanics was known for the correct picture to emerge. Today, we remember Drude because his boldly wrong model leads to the correct formula, ne 2 / m . In science, unlike politics, the good that physicists do lives after them, while the bad is oft interred with their bones. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.26 13.9 Degeneracy Pressure This material will be used in the next section, Section 13.10, but will not be needed after that. In the previous sections we have described the properties of conduction electrons in metals. Because these electrons behave as if they were nearly free, they are often referred to as a "gas" of electrons. However, the behavior of this gas is quite unlike that of an ordinary gas of molecules, such as air. The electron gas is an example of a degenerate fermi gas — called degenerate because the system is near its ground-state*, and fermi because it consists of fermions, that is, particles which obey the Pauli exclusion principle as explained in Section 10.5. One of the properties of a degenerate fermi gas is that it is extremely hard to compress. If you squeeze on a block of metal, like copper, you find that to compress it significantly requires roughly 1 million atmospheres of pressure. This incompressibility of metals is largely explained by a quantum mechanical effect called degeneracy pressure, which occurs in degenerate fermi gases. The basic physics of degeneracy pressure can be understood by considering the simplest possible fermi gas, namely a single electron, confined to a rigid 1D box, and in its ground state. The ground-state wave function is shown in Figure 13.32. If the box has length L, the wavelength of the ground state is = 2L and the ground state energy is, from (13.15) E h 2 / (8mL2 ) . If the box is compressed, decreasing L, then the kinetic energy of the ground state must increase since E proportional to 1/L2 . (Recall that in quantum mechanics, shorter wavelength corresponds to higher kinetic energy.) So compressing the container and decreasing the wavelength, necessarily increases the energy. Since the energy is increased, work must be done to perform the compression; the gas is hard to compress. The electron wants to lower its energy by spreading out over a larger volume; that is, it wants to push the walls of the enclosure outward. This pressure which the confined electron exerts on the walls of its container is called the degeneracy pressure. [FIG 13.32 new XX] This argument is readily extended to many electrons (or fermions) in a 3D box. The n quantum state of an electron in a 1D rigid box has wavelength such that th n / 2 L and its energy is p2 h2 h2 En n2 (13.18) 2m 2m 2 8 m L2 In a 3D cubical box of volume V = L3, the energy becomes * There are two distinct uses of the term "degenerate" in quantum mechanics. The first use was described in Section 8.3: Two wavefunctions are degenerate if they have the same energy. Now we encounter a different use: A many-particle system is degenerate if the system is near its ground state so that its particles are "crowded" into a limited number of single-particle states; the system is non-degenerate if the temperature is high enough that the system is far from its ground state and its particles have many empty single-particle states to choose from. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.27 h2 2 x E n 2 n y2 n z2 (13.19) 8m L where nx, ny and nz are the quantum numbers labeling the state. For many non-interacting electrons in a 3D box, the total energy is the sum of the one electron energies h2 E t ot 8 m L2 n i 2 xi n yi n zi 2 2 (13.20) where the sum is over all occupied states i. At low temperatures, each level up to the Fermi level will be occupied with 2 electrons. Now a marvelous simplification occurs. In our present analysis, only the dependence of the energy on the dimensions of the container matters. So we rewrite the total energy (13.20) as A A E t ot 2 , (13.21) L V 2/ 3 where A is a constant and V = L3 is the volume of the cubical box. We can use this simple expression for the total energy to compute the pressure of the Fermi gas. Recall from thermodynamics, that for a gas of volume V and pressure p, if the volume of a gas is changed by an amount dV, and no heat is added or removed (a so- called adiabatic change), then the work done on the gas is equal to the change in energy dW dE p dV . (13.22) The minus sign here indicates that the energy increases when the volume decreases. The pressure exerted by the gas on the walls of its container is thus dE p (adiabat ic condit ions) . (13.23) dV Combining (13.21) and (13.23), we find that the degeneracy pressure is d A 5 / 3 2 / 3 1 2 E grd p 2 AV 2 AV V , (13.24) d V V 2 / 3 3 3 3 V where, in the last step, we have used (13.21) and have assumed that the temperature is sufficiently low that the total energy Etot is approximately the ground state energy Egrd. Example 13.4. Estimate the degeneracy pressure of an electron gas with a concentration equal to that in an ordinary metal. We use equation (13.24) and our earlier estimate (13.16) of the Fermi energy, EF 4 eV. For a degenerate Fermi gas, the particles occupy single-particle states with energies that range from zero to EF. We therefore expect that the average energy per particle is roughly EF/2 (the exact answer turns out to be (3/5)EF) . The total energy of a gas is the sum of the energies of all the particles in the gas. So, if there are N particles in the gas, the total energy is E grd N E F /2, and equation (13.24) becomes 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.28 2 E grd 1 N EF 1 p n EF (13.25) 3 V 3 V 3 where n = N/V is the concentration of electrons (number/volume). For a metal, this is roughly n = 1/a3, where a 0.3 nm is a typical lattice constant, and so the pressure is EF ( 4 eV) (1.6 10-19 J / eV) p 1011 N/ m 2 106 at m. 3a 3 (3) (0.3 10-9 m)3 (13.26) (In the last step, we have used 1 atmosphere = 1 atm 105 N/m2.) This is an enormous pressure, comparable to the pressure at the center of the Earth. The reason metals do not blow apart is because this pressure is balanced by the very strong Coulomb attraction between the negative electrons and the positive ion cores. Degeneracy pressure makes degenerate fermi gases much harder to compress than ordinary molecular gases. If you squeeze on a molecular gas, such as air, at STP (standard temperature and pressure: T = 273 K, p = 1 atm) with a pressure of two atmospheres, its volume will decrease by about a factor of two This is easy to deduce from the ideal gas law; see Problem 13.xx. In contrast, to substantially compress a solid such as a block of copper requires a pressure of the same order of magnitude as the degeneracy pressure — about 106 atm. (See Problem 13.xx.). Degeneracy pressure has some remarkable consequences for the evolution of stars, as we shall see in the next section. 13.10 White Dwarfs, Neutron Stars, and Black Holes th This section, though it describes some of the most remarkable discoveries of 20 century astronomy, contains material that will not be used again and can be omitted without loss of continuity. The most spectacular examples of degenerate fermi gases occur in the realm of astrophysics. Degeneracy pressure profoundly influences the evolution of stars and leads to fantastistically dense stellar remnants called white dwarfs and neutron stars. An ordinary star, like the Sun, has a long, stable "middle age" during which the star shines because the temperature and pressure near its core are high enough for hydrogen nuclei (protons) to fuse to form helium. This fusion reaction produces the energy which makes the star shine, and it also produces an outward pressure which conteracts the inward pull of gravity, so the star has a stable size. However, eventually the supply of hydrogen near the star's core is depleted to the extent that the fusion reaction slows and the core begins to collapse because the inward pull of gravity is unopposed. Details of the subsequent evolution depend on the mass of the star, but story is usually violent. As the core collapses, it becomes hot enough that helium begins to fuse to form even heavier elements, and the outer layers of the star are blown off by the violent reactions in the core. Eventually, the nuclear fuel is exhausted and the star further collapses under its own gravity. If the star is not too much more massive than the sun, it will collapse until electron degeneracy pressure (13.24) grows large enough to counteract gravity. The star then is extremely dense, with a mass comparable to the Sun's mass packed into a sphere 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.29 the size of the Earth, as in Figure 13.33. Such a star is called a white dwarf, because, initially, it is hot enough that it is classified as "white hot". But, over time, the star cools by radiation and becomes a "black dwarf", a cold, immensely massive, cinder. Countless white dwarf stars have been identified by astronomers; they are common objects in the sky. [FIG 13.33 new XX] In 1937, a young theoretical physicist, S. Chandresakhar, discovered that if a star's mass is greater than 1.4 solar masses — the so-called Chandresakar limit — even electron degeneracy pressure is insufficient to stop to the gravitational collapse. As the star collapses into an object smaller than a white dwarf, a process called inverse beta decay occurs; at sufficiently high pressures, an electron will burrow into a proton, forming a neutron and a neutrino: p e n ν . (See Section 17.4, for discussion of closely related reactions.) The star is then composed primarily of neutrons. Neutrons, like electrons and protons, are fermions; they obey the Pauli exclusion principle and can exert degeneracy pressure just as electrons do. If the star's mass is less than about 5 solar masses, then neutron degeneracy pressure will balance gravity when the diameter of the star is about 10 km. This fantastically dense object is called a neutron star; the density at its center is about 1018 kg/m3, which is the density of "nuclear matter" in the interior of atomic nuclei. The escape speed at the surface of a neutron star is about 0.8 c, 80% of the speed of light. Problem 13.XX explore the calculation of the diameter of a neutron star. Another remarkable feature of neutron stars is their rapid rate of rotation. As a star collapses, it must spin faster, due to conservation of angular momentum — just as a skaters spins faster as they pulls in their arms. Sun-sized stars typically have rotational periods of days to months; after collapse into an object of 10 km diameter, a neutron star spins at rates of 10 to 1000 times per second. The rapid rotation of the neutron star and its magnetic field interacts with intersteller dust and gas nearby to produce a bright light source which rotates with the star, causing the star to appear to flash on and off as seen from the earth. This rapidly flashing neutron star is called a pulsar, and, since the 1960's, several dozen pulsars have been positively identified. [CHANDRESAKAR BIO] If the mass of the collapsing star is greater than roughly 5 solar masses, then even neutron degeneracy pressure is insufficient to counteract the inward pull of gravity and the star collapses into an infinitely dense, infinitesimally small object called a singularity. At present, we have no understanding of physics in the vicinity of a singularity; all our theories (relativistic quantum mechanics, general relativity) break down under such conditions. Surrounding the singularity is an imaginary sphere called the event horizon which marks the boundary at which the escape speed equals the speed of light. The event horizon around a 10-solar-mass singularity is a few kilometers in diamter. No light and no signal of any kind can escape from inside the event horizon; things can fall in, but nothing can come out. The singularity and its surrounding event horizon are therefore called a black hole. The evidence for the existence of black holes is indirect but very strong. We can "see" black holes by the effect of their gravity on their surroundings. It is believed 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.30 that super-massive black holes, of some 106 solar masses, reside at the center of most galaxies, including our own Milky Way. See Figure 13.34. [FIG 13.34 new XX] 13.11 Classical and Quantum Gases This material will be needed in the following section and in Chapter 15, Statistical Mechanics, but not elsewhere in this text. The most general definition of a "gas of particles" that one can give — a definition sufficiently broad that it is valid in the framework of either classical or quantum mechanics — is this: A gas is a collection of non-interacting or weakly-interacting particles. In this section, we look closely at the distinction between classical and quantum gases. A classical gas is one that is well described with classical mechanics, while a quantum gas is one in which quantum effects are important. We have already seen in Section 13.9 one example of a quantum gas, namely, the degenerate fermi gas formed by conduction electrons in a metal. We will see that a gas behaves classically when conditions are such that the particles of the gas are far apart, compared to their wavepacket diameters. However, for a gas of identical particles, if the particles overlap, then quantum effects cannot be ignored. The properties of a quantum gas depend crucially on whether the particles of the gas are fermions or bosons. Before describing quantum gases, let us review the properties of fermions and bosons, which were briefly discussed in Section 10.5. All particles can be classified as either fermions or bosons. A fermion is defined as a particle which obeys the Pauli exclusion principle. Examples of fermions are the electron, the proton, and the neutron. In Section 10.4, the Pauli exclusion principle was stated as: "No two electrons in a quantum system can occupy the same quantum state". We can state this principle more generally as: "No two identical fermions can occupy the same quantum state". Bosons are defined as particles which do not obey the Pauli Exclusion. In a quantum system consisting of many identical bosons, an unlimited number of the bosons can exist in the same quantum state. One example of a boson is the alpha particle, which is the nucleus of a helium atom consisting of two protons and two neutrons. Another example is the deuteron, which is a nucleus with one proton and one neutron. Note that the term "particle" is not limited to "elementary" particles such as electrons, but includes complex bound collections of particles. For instance, a neutral atom of rubidium-87, consisting of 37 protons, 37 electrons, and 50 neutrons, can be regarded as a "particle" which happens to be a boson. The deuterium atom, which is an isotope of hydrogen consisting of an electron (a fermion) bound to a deuteron (a boson), is a fermion. How does one tell if a particle is a boson or a fermion? One way, in principle, is to test whether the particle obeys the Pauli Exclusion Principle; that is, put many of the particles in a box and see if more than one can occupy a given quantum state. If only one particle can occupy each state, then the particle is a fermion; if more than one can, then it 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.31 must be a boson. Fortunately, there is an easier way to distinguish between fermions and bosons — the spin of the particle determines its type. We saw in Chapter 9 that electrons have an intrinsic angular momentum called the spin. All particles, no matter how complex, have a spin, that is, an internal total angular momentum. Like energy, the internal angular momentum of a particle or of a system of particles is always quantized; the maximum z component of the angular momentum can always be written as s , where the number s, which is called the spin, is either an integer (0, 1, 2, 3,...) or a half integer (1/2, 3/2, 5/2, etc.). Experimentally, it is found that if the total spin is an integer, then the particle is a boson; if the spin is half-integer, then the particle is a fermion. (This correspondence between spin and the Pauli principle can be derived from relativistic quantum theory.) Electrons, protons, and neutrons all have spin 1/2, so they are fermions. Composite particles, such as atoms, can be either fermions or bosons. In general, if an atom contains an even number of fermions, then it is a boson. For instance, in the ground state of the rubidium-87 atom, the spins of the 37 protons, 37 electrons, and 50 neutrons add up in such a way that the angular momenta of all (37+37+50 = 124) spin-half fermions cancel and produce a boson with spin zero. If an atom has an odd number of fermions, then its total spin is half-integer and the atom as a whole is a fermion. If a collection of identical particles is placed in a rigid box or some other quantum potential well, then which quantum states are occupied by the particles depends both on the temperature and on whether the particles are fermions or bosons. At lower temperatures, the particles tend to occupy the available states with the lowest energy, while at high temperatures, the particles tend to spread out in energy and occupy states of both low and high energy with nearly equal probability. In Figure 13.35, we compare the behavior of a gas of fermions and a gas of bosons. The figure shows the spectrum of energy levels of some quantum system, such as a rigid box or a harmonic oscillator or some other, more complex, potential. These levels are called "single-particle states" because they are the solutions of the time-independent Schrödinger equation for any one particle. If we assume that the potential energy function [U(x, y, z)] in the Schrödinger equation does not involve the spin of the particle, then the levels are degenerate; for the case of an electron, each level is at least doubly degenerate and represents two quantum states, one with spin up and one down. [FIG 13.35 new XX] We now imagine filling these quantum states with identical particles, either bosons or fermions. We assume that the particles are non-interacting, so that the presence of other particles does not affect the potential which a given particle experiences, and the solutions of the one-particle Schrödinger equation remain valid, even though there is more than one particle present. At absolute zero (T = 0 K), the particles will occupy the lowest available states. For the case of spin 1/2 fermions, the Pauli exclusion principle requires that each orbital be occupied by no more than two fermions, and so the states are filled up to a maximum level, the Fermi level. However, for bosons, the Pauli Principle does not apply; all the particles can happily occupy the same quantum state, and at T=0, the bosons all crowd into the lowest energy state, the ground state. As we have seen in previous sections, examples of degenerate fermi gases include conduction electrons in metals and in white dwarf stars, as well as neutrons in neutron stars. Degenerate fermi 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.32 gases are common. In contrast, the degenerate boson gas, also called a ―bose gas‖ or "bose-einstein condensate", is a very exotic state and is found only at extremely low temperatures. As far as we know, degenerate bose gases do not occur in nature. A human-made example of a degenerate bose gases is described in the next section. Still looking at Figure 13.35, we now imagine raising the temperature of our gas slightly. As explained in Chapter 3, at temperature T, a particle has a thermal energy of about kT, where k is Boltzmann's constant. This results in a thermal smearing of the occupancy of the levels. For fermions, some of the states above the Fermi level are now occupied, due to thermal activation of the particles, leaving some of the states below the Fermi level vacant. For bosons, not all the particles are in the ground state; some occupy excited states about one kT above the ground state. Now let us raise the temperature considerably. With the temperature high enough, then, regardless of whether the particles are fermions or bosons, they are now widely dispersed among the quantum states. The average occupancy of every state is much less than one, and it is unlikely that more than one particle will compete for occupancy of the same state. In this high temperature regime, a collection of particles behaves as a classical gas, because quantum effects due to multiple occupancy are negligible; in this classical regime, a gas of bosons behaves just like a gas of fermions. In contrast, when the temperature is sufficiently low that the occupancy of the states depends on whether the particles are fermions or bosons, then we are in the degenerate regime. In this low- temperature regime, the collection of particles is referred to as a quantum gas. The temperature required to reach the classical regime depends on the density of the particles in the system. The gas is in the classical regime if the density is low enough so that particles can "stay out of each other's way", that is, if their quantum mechnical wave packets do not overlap. We can estimate the temperature and density necessary for this classical behavior by the following argument, which has 3 main ingredients: (1) If the number of particles per volume is n, and the average distance between neighboring particles is r, then, on average, the volume per particle is r3, and 1 n . (13.27) r3 (2) A free particle with kinetic energy E and momentum p has a deBroglie wavelength related to E and p by p2 h2 E . (13.28) 2m 2m 2 (3) For a free particle at temperature T, the kinetic energy is given by E 1 m v 2 3 kT . 2 2 (13.29) where v is the mean thermal speed. This relation was discussed in Section 3.7. [FIG 13.36 new XX] 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.33 Wave packets have a size of at least the DeBroglie wavelength , so the condition of non-overlapping wavepackets can be written a ; that is, the mean distance between particles is much greater than their wavelength, as seen in Figure 13.36. We can combine points (2) and (3) [equations (13.28) and (13.29)] to obtain an expression for the deBroglie wavelength: h2 h 3 kT or . (13.30) 2m 2 2 3m k T This last expression is called the thermal DeBroglie wavelength; it is the wavelength of particles at a temperature T. At high temperatures, the DeBroglie wavelength is small, and the particle can be represented by a small, compact wave packet which behaves very much like a classical particle. At very low temperatures, however, the wavelength is large, and the wave packets describing the particles may overlap or even fill the volume of their container. Finally, we combine observation (1) [Eqn. (13.27)] with (13.30) to write the classical-gas condition a as 3/ 2 1 h 3m k T , or n 2 . (13.31) n 1/ 3 3m k T h We have achieved our goal. This last equation is gives the conditions of number density and temperature required for a gas to be in the classical regime. We can define a quantum concentration nq as 3/ 2 3m k T nq 2 (13.32) h which is the approximate particle concentration at the crossover from classical to quantum regimes. At concentrations higher than this, the wavepackets overlap strongly, quantum effects cannot be ignored, and the gas is in the degenerate quantum regime. At concentrations much less than this, the gas is in the classical regime and the ideal gas law applies. Example 13.5 What are the thermal deBroglie wavelengths of free electrons and neutrons at room temperature? At room temperature, T 300 K. From equation (13.30) we have, for electrons h 6.63 1034 6.2 109 m 6 nm 3m k T 3 9.11 10 31 1.38 10 300 23 . For neutrons, 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.34 h 6.63 1034 1.5 1011 m 0.015 nm 3m k T 3 1.67 10 27 1.38 10 300 23 . Note that for electrons, the room-temperature wavelength is much greater than the distance between atoms in a solid. Thus, for conduction electrons in a solid, the wave packets strongly overlap, and they must form a degenerate quantum gas. Example 13.6. Consider the degenerate fermi gas formed by conduction electrons in a metal. How high must the temperature be raised before the gas enters the classical regime? The crossover from degenerate fermi gas to non-degenerate classical gas occurs when the number density n is equal to the quantum concentration nq given by (13.32). Solving the equation n = nq for the temperature T, we obtain h 2 n 2/ 3 T (13.33) 3m k Since there is about 1 conduction electron per atom in a metal, the number density of conduction electrons is roughly the same as the density of atoms, and we have n = 1/a3 , where a is the lattice constant, the distance between nearest-neighbors atoms in the metal. For all solids, the lattice constant is roughly a = 0.3 nm . We now substitute n = 1/a3 into (13.33) and obtain h2 T 3m k a 2 (13.34) 6.6 10 34 2 130000 K 3 9.1 1031 1.4 10 0.3 10 23 9 2 This temperature is far above the melting point of any solid. We conclude that for any solid metal, the temperature is always well below that needed to excite its electrons into the classical regime. Conduction electrons are always in the strongly degerate quantum regime. 13.12 Bose-Einstein Condensation This material can be omitted without loss of continuity. In 1924, Einstein predicted that a gas of identical bosons will exhibit very strange behavior if cooled to a low enough temperature. Building on earlier work by the Indian physicist S. Bose (after whom bosons are named), Einstein showed that below a certain critical temperature, most of the bosons in a gas will crowd into the ground state. This 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.35 low-temperature collection of ground-state bosons is called the Bose-Einstein condensate. Before the work of Bose and Einstein, it was already well understood that the particles in a gas of bosons will all enter the ground state when the temperature is very close to absolute zero. Particles in a gas have a thermal energy of about kT, and so we expect that most of the particles will be in the ground state when the temperature is so low that this thermal energy is less than the energy of the first excited state in the system. This temperature turns out to be very, very, very low -- about 10–13 K for an ordinary gas (see Problem 13.XX). What Einstein showed was that bosons will condense into the ground state at a significantly higher temperature than this. The Bose-Einstein condensation occurs when the particles’ wavefunctions begin to overlap, namely, at the temperature of crossover from the classical to degenerate quantum regimes described in the previous section. This critical temperature, given approximately by equation (13.33), depends on the density of the gas and the mass of the particle, but for a low-pressure gas, it is typically around 10–8 K or higher (See Problem 13.XX). Einstein’s derivation is a bit too subtle to describe here, but what he showed was that the phenomenon of bose condensation is a consequence of the symmetry of the wavefunction of indistinguishable particles (described in Section 10.5). He showed that not only can bosons occupy the same state, but that below a certain critical temperature, they actually prefer to. More than 70 years passed before the 1924 prediction of Einstein was verified experimentally. In 1995, two physicists at the University of Colorado, Carl Wieman and Eric Cornell, led a team of researchers who acheived Bose-Einstein condensation (BEC) in a gas of rubidium-87 atoms. In 2001, Cornell and Wieman, along with MIT physicist Wolfgang Ketterle, received the Nobel Prize for their experimental studies of the Bose- Einstein condensation*. [Cornell and Wieman BIO] We briefly describe the original experiment of Wieman and Cornell, which is a marvel of ingenuity. The basic problem was how to cool a gas sample down to 10-7 K, a temperature that had not been achieved before. They began by suspending the gas sample in a magnetic trap so that the atoms of the gas did not touch the hot walls of the sample container. They then cooled the gas in two stages. In the first stage, a technique called laser Doppler cooling was used to cool a sample of about 1017 atoms down to 10-5 K. In laser Doppler cooling, lasers shine on the sample from six directions (Figure 13.37). The frequency of the laser light is tuned to be near an electronic transition of the atom, so that when the atom is moving toward the laser light, the Doppler-shifted frequency of the light * Yet another example of experimentalists winning a Nobel Prize for verifying a theoretical prediction due to Einstein. In 1926, French experimentalist Jean Perrin won the Prize for verifying Einstein’s theory of Brownian motion. In 1923, American experimentalist Robert Millikan won the Prize for verifying Einstein’s theory of the photoelectric effect. In 1993, American radio astronomers Russell Hulse and Joseph Taylor won the Prize for their studies of gravitational radiation from a radio pulsar, which verified Einstein’s theory of General Relativity. Einstein’s theory of Special Relativity is at the foundation of the field of High-Energy Physics, which has produced more than two dozen Prizes. Einstein won the Prize once, in 1921, for the photoelectric effect. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.36 is just right to cause an inelastic collision that slows the atom. Because of the Doppler shift, the atom only interacts with the laser beam it is heading into. In this way, the atom is slowed by one of the six beams, no matter which way it travels, and it gradually cools by transferring its energy and momentum to re-emitted light. [FIG 13.37 new XX] In the second stage of cooling, the temperature is lowered by controlled evaporation. The magnetic container is adjusted so that the fastest-moving (hottest) atoms have just enough energy to escape the sample, leaving behind cooler-than-average atoms. When the original sample of 1017 atoms was allowed to evaporate down to 2000 atoms, the temperature had dropped to 10–7 K and the Bose-Einstein transition occurred. (See Figure 13.37 and Problem 13.XX). Virtually all the rubidium atoms dropped into the ground state, and for all practical purposes, the sample was at absolute zero. This experiment was especially elegant because all the equipment was inexpensive and small enough to fit on a table-top. Cheap diode lasers and hand-wound magnetic coils were used, rather than the million-dollar, whale-sized, liquid helium cryostats and superconducting magnets found in most low-temperature laboratories. [FIG 13.38 new XX] The work of Wieman, Cornell, and Kitterle provided the first experimental verification of Bose-Einstein condensation in a gas, that is, a collection of weakly- interacting particles. However, the basic phenomena had already been observed in systems of strongly-interacting particles, the two clearest examples being superconductivity and superfluidity. Superconductivity, which is described in more detail in the next chapter, is a phenomenon which occurs in many metals. When cooled to a few degrees above absolute zero, superconducting metals loose all electrical resistance and become perfect electrical conductors. The conduction electrons in these metals interact with the lattice vibrations in a rather complex way and combine to form bound pairs of electrons that have total spin zero and are thus bosons. It is these electron- pairs that undergo a Bose-Einstein-like transition at a few degrees Kelvin and cause superconductivity. Superfluidity is a phenomenon which occurs in liquid helium. The most common isotope of helium, 4He, is a boson. Helium forms a liquid at temperatures below 4.2 K; below 2.2 K, liquid helium-4 undergoes a Bose-Einstein transition to form a so-called superfluid, which can flow through a pipe with zero friction. (See Problem 13.XX) The amazing properties of superconductors and superfluids are a consequence of a large fraction of the particles being in the quantum ground state. Although superconductors and superfluids are both fairly well understood, a detailed theoretical description has proved enormously difficult and elusive, because of complications arising from the strong interactions between particles. The original theory of Bose-Einstein condensation provides a very clear theoretical picture, but really only applies to the simple case of a gas of non-interacting particles. Now that BEC has finally been achieved in a gas, a clean comparison between theory and experiment is possible, and rapid progress is being made in this re-vitalized field. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.37 CHECK LIST FOR CHAPTER 13 CONCEPT DETAILS Types of bonding in solids Covalent, metallic, ionic, dipole-dipole Crystals and non-crystals Long-range order in crystals, short-range order only in amorphous materials. Types of crystal structures Simple cubic, FCC, BCC, diamond, etc. Metals, insulators, and semiconductors Metals have a partly-full band of electronic states. Insulators and semiconductors have a full band separated from an empty band by an energy gap. Drude model of conductivity ne 2 / m (13.11) Electronic collisions in metals Electron-impurity and electron-phonon scattering Ohm’s Law R = V/I = constant, = E/j = constant The drift velocity Average velocity of electrons in a current-carrying wire – very slow, less than 0.01 mm/s The Fermi energy Energy of conduction electrons in highest filled energy states. E F h 2 / 8mr 2 a few eV (13.16) The Fermi speed Very high intrinsic speed of conduction electrons due to Pauli exclusion principle. vF 106 m/s. Degeneracy pressure Pressure of degenerate electron gas, due to Pauli exclusion principle. p (1/3) n EF (13.25) White dwarfs, nuetron stars, black holes Stellar remanants: gravity balanced by electron degeneracy pressure, neutron degeneracy pressure, or nothing. Classical vs. quantum gases When wave packets overlap, quantum effects are important. n q 3mkT / h 2 3/ 2 Crossover at density (13.32) Bose-Einstein Condensation Bosons violate Pauli exclusion principle, with a vengence! 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.38 PROBLEMS FOR CHAPTER 13 SECTION 13.2 (BONDING OF SOLIDS) 13.1 [ = old 17.1 XX] From the data in Table 13.1 find the cohesive energies of diamond and germanium in J/mol and kcal/ mol. 13.2 [ = old 17.2 XX] From the data in Table 13.1 find the cohesive energies of NaCl and LiF in J/mol and kcal/mol. (Be careful! How many atoms are contained in a mole?) 13.3 [ = old 17.3 XX] The force between two electric dipoles (each of zero net charge) results from the imperfect cancellation of the various attractive and repulsive forces among the four charges involved. (a) Prove that the force between the two dipoles in Fig. 13.37(a) is repulsive. (b) What is it for the dipoles in Fig. 13.37(b)? 13.4 [ = old 17.4 XX] (a) Prove that the force between two dipoles with two like charges closest as in Fig. 13.37(a) is repulsive. (b) What is it if unlike charges are closest as in Fig. 13.38(b)? 13.5 [ = old 17.5 XX] The van der Waals force between two atoms arises because the fluctuating dipole moment of atom 1 induces a dipole moment in atom 2, and this induced moment is always oriented so as to experience an attractive force from atom 1. This point is illustrated in Fig. 13.2, for one orientation of the dipoles. (a) Sketch the electric field lines of dipole 1 in Fig. 13.2 and verify that the induced dipole 2 is oriented as shown. (b) Choose axes in the usual way (x across and y up the page) and put the origin at dipole 1. If atom 2 is moved onto the y axis, what is the orientation of its induced dipole moment, and what is the direction of the force on it? (c) Repeat for the case that atom 2 is put halfway between the x and y axes. SECTION 13.3 (CRYSTALS AND NONCRYSTALS) 13.6 [ = old 17.6 XX] One can think of a simple cubic lattice [Fig. 13.5(b)] as a stack of many unit cubes [Fig. 13.5(c)]. However, in counting atoms there is a danger of mis- counting, since adjacent cubes share several atoms. (a) Consider any one atom in the lattice. How many cubes share this atom? (b) Consider any one cube. How many atoms does the cube have a share of? (c) Prove that in the lattice as a whole, there is one atom per unit cube. 13.7 [ = old 17.7 XX] Consider the Na+ ion at the center of the cube shown in Fig. 13.6(b). (a) In terms of the nearest-neighbor distance r0, find the distance of any one of the next-nearest-neighbor ions. (b) How many of these ions are there? (c) Are they Na+ or Cl– ? 13.8 [ = old 17.8 XX] Consider the Na+ ion at the center of the cube shown in Fig. 13.6(b). (a) In terms of the nearest-neighbor distance r0, find the distance of any one of the third-nearest-neighbor ions. (b) How many of these ions are there? (c) Are they Na+ or Cl–? 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.39 13.9 [ = old 17.9 XX] The density of solid NaCl is 2.16 g/cm3. Find the nearest-neighbor distance. (Hint: The number density of ions is 1 ion per r03 .) 13.10 [ = old 17.10 XX] The KCl crystal has the same FCC structure as NaCl, and the nearest-neighbor distance is 0.315 nm. What is the density of KCl? (Hint: The number density of ions is 1 ion per r03 .) 13.11 [ = old 17.11 XX] In Example 13.1 [Eq. (13.2)] we estimated the potential energy of any one ion in solid NaCl. (a) Make a corresponding estimate of the PE of any one ion in KCl, whose crystal structure is the same as NaCl, but whose nearest-neighbor separation is 0.31 nm. (b) This estimate ignores the repulsive forces that come into play when the ions get very close. Thus your answer should be somewhat below the observed value of –7.2 eV. Use this observed value to find the atomic cohesive energy of KCl in eV/atom. (The ionization energy of K is 4.3 eV; the electron affinity of Cl is 3.6 eV.) 13.12 [ = old 17.12 XX] Consider the Na+ ion at the center of the cube in Fig. 13.6(b). (a) How many fourth-nearest neighbors does it have? (b) Are they Na+ or Cl– ? (c) Where are they? (d) What is their distance? 13.13 [ = old 17.13 XX] We saw in Example 13.1 that the PE of any one ion in NaCl has the form U = – ke2/r0 , where = 1.75 is called the Madelung constant. The calcula- tion of this constant in three dimensions is much more trouble than it is worth here, but the one-dimensional. Madelung constant is fairly easy to calculate, as follows: Consider a long one-dimensional "crystal" of alternating Na+ and Cl– ions spaced a distance r0 apart. Prove that the PE of any one ion is U = – ke2/r0 where = 1.39. [Hint: The series for ln(l + z) in Appendix B (with z = 1) is useful in this calculation.] 13.14 [ = old 17.14 XX] Extend the reasoning of Problem 13.6 to prove that the number of atoms per unit cell of the BCC lattice is 2. 13.15 [ = old 17.15 XX] Extend the reasoning of Problem 13.6 to prove that the number of atoms per unit cell of the FCC lattice is 4. 13.16 [ = old 17.16 XX] Gold has the FCC structure, and its density is 19.3 g/cm3. (a) Use the result of Problem 13.15 to find the edge length of the unit cube. (b) What is the nearest-neighbor separation (center to center as usual)? 13.17 [ = old 17.17 XX] Copper has the FCC structure, and the center-to-center distance between nearest-neighbor atoms is 0.255 nm. Use this information and the result of Problem 13.15 to find the density of solid copper. 13.18 [ = old 17.18 XX] Iron has the BCC structure, and its density is 7.86 g/cm3. Use the result of Problem 13.14 to find the center-to-center distance between nearest-neighbor iron atoms. 13.19 [ = old 17.19 XX] In the CsCl crystal the Cs+ ions and Cl– ions are arranged on two identical simple cubic lattices. The Cl– lattice is offset from the Cs+ lattice, so that each Cl– is at the exact center of a cube of Cs+ ions, and vice versa. Thus a unit cell of CsCl looks just like the BCC cell in Fig. 13.7 (except that the center ion is Cl–, whereas those at the corners are Cs+, or vice versa). (a) The density of CsCl is 3.97 g/cm3. Use the 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.40 result of Problem 13.14 to find the edge length of the unit cube of CsCl. (b) What is the nearest-neighbor distance? 13.20 [ = old 17.20 XX] (a) Imagine a large box full of identical rigid spheres packed in a simple cubic lattice and each touching all of its nearest neighbors. What fraction of the total volume is taken up by the spheres? (This fraction is called the packing fraction.) Answer the same question for (b) FCC and (c) BCC packing. (d) Which arrangement packs the most spheres per volume? 13.21 [ = old 17.21 XX] To estimate the tensile strength of the material in Example 13.2, we had to know the force needed to pull two atoms apart. To get an estimate of this force consider the semiclassical model of the H2 molecule in Fig. 16.34 (Problem 16.11). Using the result s = R 0 / 3 from Problem 16.11 and taking R0 = 0.3 nm as a representative lattice spacing, find the total force of one atom on the other. 13.22 [ = old 17.22 XX] Problem 13.21 suggests one way to estimate the force needed to separate two bonded atoms; here is another. The graph in Fig. 16.6 shows the energy of an Na+- Cl– pair as a function of their separation. Since the slope of this graph is the force between the ions, the maximum force needed to pull them apart is just the maximum slope of the graph. Use the graph to show that this force is of order 3 10–9 newton. 13.23 [ = old 17.23 XX] In Example 13. 1, Eq. (13. 1), we wrote down the first three terms in the electrostatic PE of an ion in the NaCl lattice. What are the next two terms? 13.24 [ = old 17.24 XX] In Example 13. 1, Eq. (13. 1), we said that the electrostatic PE of any one ion in NaCl is Ues = – ke2/r, where = 1.75. This is the electrostatic energy of simple point charges, and we should have included a contribution due to the repulsion of the ions at close range. This repulsive contribution can be approximated by a term of the form Urep = + a/r n, where a and n are positive constants. Thus the total PE is ke 2 U (r ) U es (r ) U rep (r ) n r r (a) At the equilibrium separation r = r0 , this energy is a minimum. Use this fact to find the constant a, and hence show that the total PE of an ion at equilibrium is U (r0 ) ke 2 1 1 r0 n (b) This result differs from our rough estimate (13.2) only by the second term 1/n. For NaCl, r0 = 0.281 nm and n = 8. Use these values to find the total PE of an ion in NaCl. Compare with the rough answer of –9.0 eV found in Example 13.1 and the observed value of – 7.92 eV. SECTION 13.4 (ENERGY LEVELS OF ELECTRONS IN SOLIDS; BANDS) 13.25 [ = old 17.25 XX] The density of solid silicon is 2.33 g/cm3. (a) Estimate the total number, N, of silicon atoms in a microcrystal 1 m 1 m 1 m. (b) The valence band in silicon is about 1 eV wide. Given that there are 4N states in this band, make a rough 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.41 estimate of the average separation between states in the valence band of this crystal. (c) Consider a silicon solar cell that is 1 cm 1 cm. 1 mm; repeat both calculations for these dimensions. 13.26 [ = old 17.26 XX] The density of solid germanium is 5.32 g/cm3. (a) Estimate the total number, N, of atoms in a microcrystal 1 m 1 m 1 m. (b) The conduction band in germanium is about 2 eV wide. Given that there are 4N states in this band, make a rough estimate of the average separation between states in the conduction band of this crystal. (c) Consider a germanium crystal that is 1 cm 1 cm 1 cm; repeat both calculations for these dimensions. 13.27 [ = old 17.27 XX] Imagine a crystal containing just one type of atom on a simple cubic lattice. If the lattice spacing is 0.2 nm and the whole crystal is 3 mm 3 mm 3 mm, how many states are there in the 2p band? 13.28 [ = old 17.28 XX] Consider a solid containing N atoms and a band that evolved from an atomic level of angular momentum 1. How many states does this band contain? Check that your answer gives 2N for an s band and 6N for a p band. SECTIONS 13.5 AND 13.6 (CONDUCTORS AND INSULATORS) 13.29 [ = old 17.29 XX] Consider a single crystal of an insulator with a band gap of 3.5 eV. What colors of visible light will this crystal transmit? 13.30 [ = old 17.30 XX] Crystalline sulfur absorbs blue light but no other colors; as a result, it is a pale yellow transparent solid. It is also an electrical insulator. From this information deduce the width of the band gap of crystalline sulfur, and describe the occupancy of the bands above and below the gap. 13.31 [ = old 17.31 XX] Germanium is a semiconductor with a band gap of 0.8 eV. What colors of visible light does it transmit? 13.32 [ = old 17.32 XX] Explain why many insulators are transparent to visible light, whereas most semiconductors are opaque to visible light but transparent to infrared. 13.33 [ = old 17.33 XX] The longest wavelength of electromagnetic radiation that is absorbed in gallium arsenide is 890 nm. What is the band gap of this semiconductor? 13.34 [ = old 17.38 XX] Consider a hypothetical element that forms a solid with bands as shown in Fig. 13.38. (a) Suppose that the isolated atom has configuration1s22s2. If its equilibrium separation is r0 = a, is the solid a conductor or insulator? What if r0 = b? Answer both questions for the case that the atomic configuration is (b) 1s22s22p1 and (c) 1s22s22p6. SECTION 13.6 (DRUDE MODEL OF CONDUCTIVITY) 13.35 [ = new XX] (a) Make a sketch of current I vs. voltage V for an ohmic resistor. What is the significance of the slope of your graph? (b) When a 120V light bulb is on, the temperature of the filament is about 3000 C, and the resistance due to lattice 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.42 vibrations is approximately 10 times the resistance of the filament at room temperature. Make a sketch of the current I vs. voltage V for a tungsten light bulb filament over a range of voltages from 0 to 120 V. (c) Why is a light bulb most likely to burn out, immediately after it is first turned on? 13.36 [ = new XX] (a) Consider a simple cubic lattice of atoms, in which the number density of atoms is na . Show that the distance d between nearest neighbors is given by d = na -1/3. (b) In a metal, we can define a distance rs as the radius of a sphere whose volume is the volume per conduction electron. Show that rs is related to the conduction electron density n by the formula rs 3 / ( 4 n )1/ 3 (c) What is the relation between rs and d for a monovalent metal with a simple cubic lattice? 13.37 [ = new XX] Compute the number density n (number per volume) of conduction electrons in silver, assuming one conduction electron per atom. The mass density of silver is 10.5 grams/cm3. 13.38 [ = new XX] Consider a metal in which the distance between nearest-neighbor atoms is d = 0.3 nm. (a) Assuming a simple cubic lattice structure, and one conduction electron per atom, what is the conduction electron density n (number per volume)? (b) Assume that a 1 mm diameter wire made of this metal carries a current of 50 A (a rather large current). What is the current density j? (c) What is the drift speed in this current- carrying wire? 13.39 [ = new XX] In Section 13.8, we show that the mean speed of conduction electrons in a metal is roughly 0.003 c (this is approximately the Fermi speed). Suppose that the velocity of a particular conduction electron is known to within 1%, so that the uncertainty in the speed is approximately v = 3 10–5 c. Use the Heisenberg uncertainty principle to show that the resulting uncertainty in the position of the electron is greater than, but not enormously greater than, the distance between atoms in a solid. (Since this calculation is only approximate, you can assume 1D motion.) This calculation is the justification for the use of the semi-classical approximation of the Drude model, in which we assume that quantum mechanical electrons have a well-defined position and momentum. 13.40 [ = new XX] Consider a cylindrical resister of length L, cross-sectional area A, resistivity , and resistance R. The resistor has a voltage difference V between its ends and carries a current I. Show that ―Ohm’s Law‖ V = IR is equavalent to E = j , with resistance R related to the resistivity by the R = L/A. 13.41 [ = new XX] One can define the scattering time in the following way: the probablity P that an electron suffers a collision during an infinitesimal time interval dt is P (collision in t ime dt ) dt / . (13.35) Note that the probability that the electron suffers no collision during a time interval dt is then given by P (no collision in dt ) 1 dt / . (13.36) 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.43 Consider now the probability P (t ) that a given electron undergoes no collision during a finite time interval t. (a) Show that P’ obeys the equation dP dt . (13.37) P [Hint: First, argue that P (t dt ) P (t )(1 dt / ) .] (b) Using (13.37), solve for P (t ) and sketch this function. (c) Suppose that a collision occurs at time t = 0. Argue that the probability P (t ) that the next collision occurs between time t and t + dt is given by dt P (t ) dt e t / . (13.38) (d) Use (13.38), to compute the average time, t , between collisions. (e) Compute the rms average time t 2 between collisions. SECTIONS 13.7 and 13.8 (ELECTRONIC COLLISIONS IN METALS and THE FERMI SPEED) 13.42 [ = new XX] Fig.13.29 shows the dependence of resistivity on temperature for a pure metal and a semiconductor. Sketch a new graph, showing the dependence of conductivity on temperature for a metal and a semiconductor. Be sure to label your curves clearly. 13.43 [ = new XX] At high temperatures (T 500 K), the resistivity of tungsten is approximately proportional to the absolute temperature ( T). A hot tungsten light bulb filament maintains a stable temperature because the electrical power into the bulb (P = I 2R = V 2/R) is exactly balanced by the power out due to cooling by radiation and conduction. At high temperatures, the cooling is dominated by radiation, and the power radiated is given by the formula , P = A T 4, where A is the surface area of the body, and both and are constants, called the emissivity and the Stephan-Boltzmann constant*, respectively. (a) Show that the temperature of a hot light bulb filament with voltage V is proportional to V 2/5. (b) Show that the temperature of a hot filament carrying a current I is proportional to I 2/3. 13.44 [ = new XX] Estimate the conductivity of an alloy or very dirty metal. Assume a simple cubic lattice, one conduction electron per atom, and a mean free path of about 2 lattice constants (twice the distance between atoms). Take the Fermi speed to be Be aware: the Greek letter (sigma) is used to denote both the conductivity and the Stephan- * Boltzmann constant. Don’t confuse them. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.44 0.003 c, typical of metals. How does this compare to the conductivity of copper at room temperature? 13.45 [ = new XX] At room temperature, the mean free path in a clean metal sample is about 150nm. The ―triple-R‖ of the metal [R(300K)/R(4K) ] is measured to be 100. What is the mean free path in the metal at T = 4 K? 13.46 [ = new XX] Repeat the calculation leading to equation (13.16)for the Fermi energy for the case of a divalent metal (one with two conduction electrons per atom). (Hint: For a simple cubic lattice, the number density n is related to the nearest-neighbor separation r by n = 1/r3.) 13.47 [ = new XX] Estimate the mean free path in a metal with a resistivity of 5 cm (typical of a good metal at room temperature). Assume a typical metal: one conduction electron per atom, a nearest neighbor atomic separation of 0.3 nm, and a Fermi speed of 106 m/s. 13.48 [ = new XX] (a) Using Drude’s formula = ne2 / m and the known values of the conductivity of metals, make an order-of-magnitude estimate of the scattering time of . (b) Compute the thermal speed of an electron at room temperature. Drude assumed, incorrectly, that the mean speed of conduction electrons is given by their thermal speed. He also assumed, incorrectly, that the atomic nuclei in the metals scatter conduction electrons, which implies a mean free path of a few lattice constants. (c) Use the thermal speed and the scattering time to compute the mean free path. Your answer should be a distance equal to a few lattice constants, consistent with Drude’s (incorrect) assumption of atomic scattering. This calculation shows that, although Drude’s made two incorrect assumptions, his model was self-consistent. SECTION 13.9 (DEGENERACY PRESSURE ) 13.49 [ = new XX] An particle in a 1D rigid box of length L (an infinite square well) is in it ground state. The box is compressed to length L/2. (a) By what factor does the ground-state energy increase? (b) By what factor does the first excited state energy increase? (c) Repeat the calculations in (a) and (b) for the case of a particle in a 3D box of volume V = L 3. 13.50 [ = new XX] Eqn (13.16) is an approximate expression for the Fermi energy in terms of the mean distance r between nearest neighbor electrons (which is the same as the distance between nearest-neighbor atoms in a monovalent metal). (a) Rewrite this expression to give the Fermi energy as a function of the number density of electrons n (number per volume). (b) Consider a metal with a Fermi energy of 4eV. What is the Fermi energy if this metal is compressed to twice its normal density? (As the next problem shows, such compression requires enormous pressure.) Hint: you should be able to compute the answer without looking up Planck’s constant h or the mass of the election. 13.51 [ = new XX] (a) An ideal gas at standard temperature and pressure (STP: T = 273K, p = 1 atm = 1.01 × 105 Pa) is compressed to one-half of its original volume while the temperature is held constant. What is the pressure of the compressed gas? Give the 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.45 answer in atmospheres. (b) Why is the degeneracy pressure unimportant in an ordinary gas (such as air) at STP? (c) A metal with a Fermi energy of 4eV and an electron degeneracy pressure of 106 atm is compressed to one-half of its initial volume. What is the new value of the degeneracy pressure? Hint: use the results of the previous problem. SECTION 13.10 (WHITE DWARFS, NEUTRON STARS, AND BLACK HOLES ) 13.52 [ = new XX] A neutron star consists of neutrons packed so closely together that they touch, just as in the interior of an atomic nucleus. Estimate the density of a neutron star, given that the radius of a neutron is about 1 fm (10-15 m). 13.53 [ = new XX] Consider a star with a mass 3 times that of the Sun, a rotational period of 30 days, and an average density equal to that of water (which is a typical density for a star) . If the star collapses into a neutron star with a diameter of 10 km, what is the new rotational period? Assume that the original star and the final neutron star both have uniform densities. Hint: use conservation of angular momentum. (This calculation produces a higher rotational rate than is usually found in neutron stars. Our calculation is flawed because normal stars, like the Sun, do not have a uniform density. Instead, most of the mass is concentrated near the center.) 13.54 [ = new XX] (a) Use non-relativistic classical mechanics to derive an expression for the radius of the event horizon surrounding a black hole of mass M. (In this case, a non-relativistic calculation happens to produce exactly the right answer.) (b) What is the event horizon radius of a 10 solar mass black hole? (c) What is the acceleration of gravity at the event horizon of a 10 solar mass black hole? 13.55 [ = new XX] Derive an expression for the acceleration of gravity at the event horizon surrounding a black hole of mass M. Curiously, the accelation of gravity at the event horizon decreases as the mass of the black hole increases. How massive is a black hole which has an acceleration of 1 g (9.8 m/s2) at its event horizon? Give your answer in solar masses. 13.56 [ = new XX] Estimation of the radius of a neutron star. (a) Estimate the gravitational pressure at the center of a star of mass M and radius R, assuming a uniform mass density. A fairly accurate estimate can be made from the following rough calculation: consider the star to be made up of 2 hemispheres each of mass M/2. The centers of the two hemispheres are roughly a distance R apart. The gravitational force between the hemispheres is then approximately F = G (M/2)2/R 2. This force is spread out over the area of contact A between the hemispheres and the pressure at the center is roughly p = F/A. (b) Starting with equations (13.16) and (13.25), derive an expression for the neutron degeneracy pressure in terms of R and M. Hint: the mass density of the star, the number density n of neutrons, and the neutron mass m are related by = m n. (c) The neutron star has a stable size because the inward gravitational pressure is equal to the outward degeneracy pressure. Derive an expression for the radius of a neutron star of mass M. If your expression is correct, it should show that the radius R and the mass M are related by R M –1/3. A larger mass produces a smaller radius! (d) Calculate the radius of a 2 solar mass neutron star. 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc New Chapter 13, Solids I: Theory 13.46 SECTION 13.12 (BOSE-EINSTEIN CONDENSATION ) 13.57 [ = new XX] Separation of energy levels in a gas. (a) Review Sections( 7.4 and 8.3) and then compute the separation between the ground state energy and the first excited state energy for a particle in a rigid cubical box of edge length L. (b) Consider a gas of non-interacting bosons with mass m equal to that of helium atoms, contained in a cubical box, with edge length L = 1 cm. If no Bose-Einstein condensation were to occur, how low would the temperature have to be for most of the atoms to be in the ground state? 13.58 [ = new XX] Bose-Einstein Condensation in rubidium vapor. Use equation (13.33) to estimate the temperature at which a gas of 2000 rubidium-87 atoms confined to a volume of 10-15 m3 will undergo a Bose-Einstein transition. These were the conditions in the original experiment of Cornell and Wieman. 13.59 [ = new XX] Critical temperature for superfluid He-4. Liquid 4He undergoes a transition to a superfluid at a critical temperature Tc , which should be given approximately by (13.33). [Equation (13.33) was derived for the case of a gas of weakly- interacting particles and is not expected to be very accurate for a system of strongly- interacting particles such as liquid helium.] Use (13.33) to get an order-of-magnitude estimate of the temperature at which liquid helium undergoes a Bose-Einstein transition and forms a superfluid. The mass density of liquid helium of 0.15 grams/cm3. (The observed value fo the critical temperature of liquid helium is Tc = 2.17 K.) 11/22/11 4:08 PM 2e48dfc1-c601-42f3-8c55-aeafe92be176.doc